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Binomial thresholds and unknown parameters

Learning goals

  • Expresses a binomial probability requirement as an equation or inequality.
  • Finds a cutoff or count interval meeting a probability requirement.
  • Determines the trial count from binomial distribution information.
  • Determines an unknown success probability from distribution information.
  • Checks adjacent integer candidates and the required direction of an inequality.
  • Uses a normal approximation to estimate a binomial coverage interval when justified.

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Binomial thresholds and unknown parameters 1
A binomial random variable has the distribution below. <table> <tr><th>\(x\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{27}{64}\)</td><td>\(\frac{27}{64}\)</td><td>\(\frac{9}{64}\)</td><td>\(\frac{1}{64}\)</td></tr> </table> Find the smallest whole number \(k\) for which \(P(X\ge k)\le0.20\).

Hints

- Read each tail as the sum of probabilities from its cutoff upward. - Compare successive cutoffs with the stated probability limit. - Confirm that the preceding cutoff does not meet the limit.

Solution

1. For \(k=1\), the probability is \(\frac{27+9+1}{64}=\frac{37}{64}>0.20\). 2. For \(k=2\), the probability is \(\frac{9+1}{64}=\frac{10}{64}=0.15625\le0.20\). 3. No smaller whole-number cutoff qualifies, so \(k=2\).

Answer

\(k=2\).
Binomial thresholds and unknown parameters 2
A pharmaceutical company packages tablets in blister packs of \(20\). Suppose each tablet has the same probability \(p\) of being defective, independently of the others. The probability that a randomly selected blister pack contains at least one defective tablet is exactly \(5\%\). Find \(p\). Round your answer to four decimal places.

Hints

- Express “at least one defective tablet” using a complement. - What is the probability that all \(20\) tablets are not defective? - Set up an equation involving the probability of zero defective tablets and solve for \(p\). - Use a twentieth root when isolating \(1-p\).

Solution

1. Let \(X\) be the number of defective tablets in a pack. Then \(X\) has a binomial distribution with \(n=20\) and defect probability \(p\). 2. The condition is \(P(X\ge 1)=1-P(X=0)=0.05\). 3. Therefore, \(1-(1-p)^{20}=0.05\), so \((1-p)^{20}=0.95\). 4. Solve for \(p\): \(p=1-(0.95)^{1/20}\). 5. Numerically, \(p\approx 0.0025614\), which rounds to \(0.0026\).

Answer

The probability that one tablet is defective is approximately \(0.0026\), or \(0.26\%\).
Binomial thresholds and unknown parameters 3
A spinner is divided into \(5\) equal sections, and exactly \(1\) section is red. The spinner is spun independently \(8\) times. a) Find the probability of exactly \(3\) red results. Give an exact binomial expression and a decimal approximation to four decimal places. b) Find the probability of at least \(2\) red results. Give an exact complement expression and a decimal approximation to four decimal places. c) What is the minimum number of spins needed so that the probability of landing on red at least once is greater than \(90\%\)?

Hints

- Use \(p=1/5\) for red. - For part b), use the complement of \(0\) or \(1\) red. - For part c), use the complement of no red results.

Solution

1. The red probability is \(p=0.2\). 2. \(P(X=3)=\binom{8}{3}(0.2)^3(0.8)^5\approx0.1468\). 3. \(P(X\ge2)=1-\left[(0.8)^8+8(0.2)(0.8)^7\right]\approx0.4967\). 4. For part c), \(1-(0.8)^n>0.90\), so \(n>10.319\ldots\). The minimum integer is \(11\).

Answer

a) \(\binom{8}{3}(0.2)^3(0.8)^5\approx0.1468\). b) \(1-[(0.8)^8+8(0.2)(0.8)^7]\approx0.4967\). c) \(11\) spins.
Binomial thresholds and unknown parameters 4
A multiple-choice vocabulary quiz has \(10\) questions. Each question has \(4\) answer choices, exactly one of which is correct. Aiko guesses independently and uniformly at random on every question. a) Find the probability of exactly \(3\) correct answers. Give an exact binomial expression and a decimal approximation to four decimal places. b) Find the probability of at least \(1\) correct answer. Give an exact complement expression and a decimal approximation to four decimal places. c) What should the minimum passing score be so that the probability of passing by guessing alone is less than \(15\%\)?

Hints

- The success probability is \(1/4\). - Use a complement for at least one correct answer. - Test upper-tail probabilities for candidate passing scores.

Solution

1. \(X\sim\operatorname{Bin}(10,0.25)\). 2. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). 3. \(P(X\ge1)=1-(0.75)^{10}\approx0.9437\). 4. \(P(X\ge4)\approx0.2241>0.15\), while \(P(X\ge5)\approx0.0781<0.15\). 5. Therefore, the minimum passing score is \(5\).

Answer

a) \(\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). b) \(1-(0.75)^{10}\approx0.9437\). c) \(5\) correct answers.
Binomial thresholds and unknown parameters 5
A hostel has \(45\) beds. For one weekend, it accepts \(52\) reservations because historical data show that each guest with a reservation arrives with probability \(0.80\). Assume the guests' arrival decisions are independent. a) What is the probability that more guests arrive than there are beds? Round to four decimal places. b) The hostel wants the probability of overbooking—more than \(45\) guests arriving—to be at most \(10\%\). What is the maximum number of reservations it may accept?

Hints

- Model the number of arriving guests with a binomial distribution. - Overbooking occurs when the arrival count exceeds \(45\). - For the maximum in part b), compare consecutive reservation counts around the threshold.

Solution

1. With \(52\) reservations, let \(X\sim\operatorname{Bin}(52,0.80)\). 2. Overbooking means \(X>45\), so \(P(X\ge46)\approx0.0823\). 3. With \(52\) reservations, the risk \(0.0823\) is at most \(0.10\); with \(53\) reservations, the risk is about \(0.1420\), above the limit. 4. Adding a reservation cannot reduce the possible arrival count. Because overbooking risk is nondecreasing as reservations increase, \(52\) is the maximum permitted number.

Answer

a) \(P(X>45)\approx0.0823\). b) \(52\) reservations.
Binomial thresholds and unknown parameters 6
The random variable \(X\) has a binomial distribution with \(n=10\). The mean \(E(X)\) is known to be an integer. The table shows part of the probability distribution. <table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.3487\)</td> <td>\(0.3874\)</td> <td>\(0.1937\)</td> <td>\(0.0574\)</td> <td>\(0.0112\)</td> </tr> </table> a) Determine the success probability \(p\). b) Find \(P(1\le X\le3)\). c) Find \(P(X\ge1)\).

Hints

- Use the fact that \(E(X)=np\) is an integer to list the possible values of \(p\). - Use the row entry for \(P(X=0)\) to identify which possible value of \(p\) fits. - Add the table entries that correspond to the requested interval. - For “at least one,” consider the complement.

Solution

1. Because \(E(X)=np=10p\) is an integer, \(p\) must be one of \(0,0.1,0.2,\ldots,1\). 2. The table gives \(P(X=0)=(1-p)^{10}\approx0.3487\). Testing the possible values shows that \(p=0.1\), since \((0.9)^{10}\approx0.3487\). 3. Add the listed probabilities: \(P(1\le X\le3)=0.3874+0.1937+0.0574=0.6385\). 4. Use the complement: \(P(X\ge1)=1-P(X=0)=1-0.3487=0.6513\).

Answer

a) \(p=0.1\) b) \(P(1\le X\le3)=0.6385\) c) \(P(X\ge1)=0.6513\)
Binomial thresholds and unknown parameters 7
A multiple-choice test has \(12\) questions. Each question has four answer choices, exactly one of which is correct. Noura guesses randomly and independently on every question. Let \(X\) be the number of correct answers. a) Find the probability that Noura answers at least \(5\) questions correctly. Round to four decimal places. b) Find the smallest integer \(k\) for which the probability of getting at least \(k\) correct answers by guessing is less than \(10\%\).

Hints

- Use \(p=1/4\) for a correct guess. - Part a is an upper-tail probability. - For part b, compare consecutive integer cutoffs around the \(10\%\) threshold.

Solution

1. \(X\sim\operatorname{Bin}(12,0.25)\). 2. \(P(X\ge5)=\sum_{j=5}^{12}\binom{12}{j}(0.25)^j(0.75)^{12-j}\approx0.1576\). 3. For the cutoff, compare successive upper-tail probabilities. 4. \(P(X\ge5)\approx0.1576>0.10\), while \(P(X\ge6)\approx0.0544<0.10\). 5. Therefore, the smallest qualifying value is \(k=6\).

Answer

a) \(P(X\ge5)\approx0.1576\). b) \(k=6\).
Binomial thresholds and unknown parameters 8
A container holds \(20\) marbles. Some are blue, and the rest are yellow. A marble is drawn \(8\) times with replacement. The probability of drawing at most \(2\) blue marbles is approximately \(67.85\%\). How many blue marbles are in the container?

Hints

- Which probability distribution applies when the draws have two outcomes and are made with replacement? - How is the probability of drawing blue related to the number of blue marbles? - Write a cumulative probability for drawing at most \(2\) blue marbles. - Test possible whole-number counts of blue marbles and compare the result with the given probability.

Solution

1. Let \(k\) be the number of blue marbles. Then the probability of drawing blue on one draw is \(p=\frac{k}{20}\). 2. Let \(X\) be the number of blue marbles drawn in \(8\) draws. Because the draws are made with replacement, \(X\) has a binomial distribution with \(n=8\) and \(p=\frac{k}{20}\). 3. The condition is \(P(X\le 2)\approx 0.6785\). 4. Test possible integer values of \(k\). For \(k=4\), \(p=0.20\) and \(P(X\le 2)\approx 0.7969\). 5. For \(k=5\), \(p=0.25\) and \(P(X\le 2)\approx 0.6785\). 6. As \(k\) increases, \(P(X\le 2)\) decreases, so the matching value is unique. Therefore, the container has \(5\) blue marbles.

Answer

There are \(5\) blue marbles in the container.
Binomial thresholds and unknown parameters 9
A shipment contains \(25\) electronic components. For quality control, a component is selected, checked, and returned to the shipment before the next selection. This is repeated \(10\) times. The probability of selecting at least \(1\) defective component is approximately \(89.26\%\). How many defective components are in the shipment?

Hints

- Use the complement of selecting at least \(1\) defective component. - What is the probability of selecting no defective components in all \(10\) draws? - Express the one-draw defect probability as a fraction of the \(25\) components. - Solve the resulting equation for the unknown probability.

Solution

1. Let \(k\) be the number of defective components. Then the probability of selecting a defective component on one draw is \(p=\frac{k}{25}\). 2. Let \(X\) be the number of defective selections in \(10\) draws. Because each component is returned, \(X\) has a binomial distribution with \(n=10\) and \(p=\frac{k}{25}\). 3. Using the complement, \(P(X\ge 1)=1-P(X=0)=1-(1-p)^{10}\). 4. Set \(1-(1-p)^{10}\approx 0.8926\). Then \((1-p)^{10}\approx 0.1074\). 5. Taking the tenth root gives \(1-p\approx 0.8\), so \(p\approx 0.2\). 6. Therefore, \(k=25\cdot0.2=5\).

Answer

The shipment contains \(5\) defective components.
Binomial thresholds and unknown parameters 10
Assume a particular side effect occurs in \(5\%\) of patients, independently from patient to patient. Let \(X\) be the number of patients who experience the side effect in a group of \(n\) patients. a) Find the sample size \(n\) for which the mean number of patients with the side effect is exactly \(20\). b) For this sample size, find the standard deviation \(\sigma\). c) Check whether a normal approximation is appropriate using the success-failure condition \(np\ge10\) and \(n(1-p)\ge10\).

Hints

- How does the binomial mean depend on the number of patients? - Once you know the sample size, which expression measures the spread of the count? - Compare the two expected outcome counts with the thresholds in part c).

Solution

1. Since \(E(X)=np=20\) and \(p=0.05\), \(n=\frac{20}{0.05}=400\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot0.05\cdot0.95}=\sqrt{19}\approx4.36\). 3. The success-failure values are \(np=400\cdot0.05=20\) and \(n(1-p)=400\cdot0.95=380\). 4. Both are at least \(10\), so the stated normal-approximation condition is met.

Answer

a) \(n=400\). b) \(\sigma=\sqrt{19}\approx4.36\). c) Yes; \(np=20\) and \(n(1-p)=380\) are both at least \(10\).
Binomial thresholds and unknown parameters 11
Let \(X\sim\operatorname{Bin}(60,\frac{1}{3})\). a) Find the probability that \(X\) differs from its mean by at most \(2\). Round to four decimal places. b) Find \(P(|X-20|\ge4)\), rounded to four decimal places. c) Find the smallest nonnegative integer \(k\) for which \(P(|X-20|\le k)>0.5\).

Hints

- Find the binomial mean first. - Translate each absolute-value condition into integer bounds. - For part c, test small integer radii in increasing order.

Solution

1. The mean is \(60\left(\frac{1}{3}\right)=20\). 2. Part a is \(P(18\le X\le22)\approx0.5061\). 3. Part b is \(P(X\le16)+P(X\ge24)\approx0.3377\). 4. For part c, \(P(|X-20|\le1)<0.5\), while \(P(|X-20|\le2)\approx0.5061>0.5\). 5. Thus, the smallest value is \(k=2\).

Answer

a) \(0.5061\). b) \(0.3377\). c) \(k=2\).
Binomial thresholds and unknown parameters 12
Let \(X\) be a binomial random variable with \(n=30\) and \(p=0.6\). Use a calculator or spreadsheet to determine: a) every value \(k\in\{0, 1, \ldots, 30\}\) for which \(P(X=k)\ge0.1\); b) every value \(k\in\{0, 1, \ldots, 30\}\) for which \(P(X=k)<0.05\).

Hints

- Expect the largest probabilities near the mean. - Use a table feature to view many individual probabilities efficiently. - Distinguish carefully between \(\ge\) and \(<\). - Check the probabilities immediately inside and outside each boundary.

Solution

1. For each \(k\), calculate \(P(X=k)=\binom{30}{k}(0.6)^k(0.4)^{30-k}\). The probabilities are largest near the mean \(np=18\). 2. Around the first cutoff, \(P(X=15)\approx0.0783\), \(P(X=16)\approx0.1101\), \(P(X=20)\approx0.1152\), and \(P(X=21)\approx0.0823\). Therefore, \(P(X=k)\ge0.1\) for \(k\in\{16, 17, 18, 19, 20\}\). 3. Around the second cutoff, \(P(X=14)\approx0.0489\), \(P(X=15)\approx0.0783\), \(P(X=22)\approx0.0505\), and \(P(X=23)\approx0.0263\). 4. Therefore, \(P(X=k)<0.05\) for \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\).

Answer

a) \(k\in\{16, 17, 18, 19, 20\}\) b) \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\)
Binomial thresholds and unknown parameters 13
A binomial random variable \(X\) has parameters \(n=100\) and \(p=0.25\). Use appropriate technology to find: a) the smallest nonnegative integer \(k\) such that \(P(X\le k)\ge 0.95\); b) the smallest nonnegative integer \(k\) such that \(P(X>k)\le 0.01\).

Hints

- Use a binomial cumulative distribution function to evaluate probabilities of the form \(P(X\le k)\). - Rewrite the condition in part b) using a complement. - The requested value is the first integer at which the inequality becomes true. - Check both the proposed value and the preceding integer.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.25)\). 2. For part a), \(P(X\le 31)\approx 0.9307<0.95\), while \(P(X\le 32)\approx 0.9554\ge 0.95\). Therefore, the smallest value is \(k=32\). 3. For part b), rewrite the condition as \(P(X\le k)\ge 0.99\). 4. Since \(P(X\le 34)\approx 0.9836<0.99\) and \(P(X\le 35)\approx 0.9906\ge 0.99\), the smallest value is \(k=35\).

Answer

a) \(k=32\) b) \(k=35\)
Binomial thresholds and unknown parameters 14
At a tour company's information booth, \(20\%\) of visitors are interested in an international trip. During one hour, \(80\) visitors stop at the booth. Assume their interests are independent, and let \(X\) be the number interested in an international trip. a) What is the minimum number of international-travel brochures the company should have so that the brochures are sufficient for all interested visitors with probability at least \(95\%\)? b) Suppose the company has \(18\) brochures. Find the probability that at least one brochure remains at the end of the hour. Round to four decimal places.

Hints

- Translate “enough with probability at least \(95\%\)” into a cumulative-probability condition. - Compare consecutive integer brochure counts to prove minimality. - Determine exactly when at least one of \(18\) brochures remains.

Solution

1. \(X\sim\operatorname{Bin}(80,0.20)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.95\). 3. Since \(P(X\le21)\approx0.9340<0.95\) and \(P(X\le22)\approx0.9612\ge0.95\), the minimum is \(22\) brochures. 4. With \(18\) brochures, at least one remains exactly when fewer than \(18\) visitors want one, so \(P(X\le17)\approx0.6708.\)

Answer

a) \(22\) brochures. b) \(P(X\le17)\approx0.6708\).
Binomial thresholds and unknown parameters 15
A home improvement store runs a promotion in which each customer rolls a fair six-sided die at checkout. A customer who rolls a \(6\) receives an instant-discount coupon. The die is rolled independently \(150\) times during the morning. a) What is the minimum number of coupons the store should prepare so that there are enough for all winners with probability at least \(90\%\)? b) The manager prepares \(30\) coupons. Find the probability that more customers win than there are coupons available. Round to four decimal places.

Hints

- Use \(p=1/6\) for a winner. - Translate the required confidence level into a cumulative-probability condition. - For part b), use the complement of having at most \(30\) winners.

Solution

1. Let \(X\sim\operatorname{Bin}(150,\frac16)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.90\). 3. Since \(P(X\le30)\approx0.8842<0.90\) and \(P(X\le31)\approx0.9197\ge0.90\), the minimum is \(31\) coupons. 4. With \(30\) coupons, a shortage occurs when \(X>30\), so \(P(X>30)=1-P(X\le30)\approx0.1158.\)

Answer

a) \(31\) coupons. b) \(P(X>30)\approx0.1158\).
Binomial thresholds and unknown parameters 16
A multiple-choice test has \(5\) answer choices for each question, exactly one of which is correct. A test taker guesses independently and randomly on every question. Find the minimum number of questions needed so that the probability of answering at least \(1\) question correctly is: a) at least \(80\%\). b) at least \(95\%\).

Hints

- What is the probability of guessing one question correctly? - Express the probability of answering no questions correctly. - Use the complement for at least \(1\) correct answer. - Use logarithms to solve for an unknown exponent. - The number of questions must be a whole number.

Solution

1. The probability of guessing one question correctly is \(p=\frac{1}{5}=0.2\), so the probability of an incorrect answer is \(0.8\). 2. For \(n\) questions, \(P(X\ge 1)=1-(0.8)^n\). 3. For part a), require \(1-(0.8)^n\ge 0.80\), so \((0.8)^n\le 0.20\). 4. Thus, \(n\ge \frac{\ln(0.20)}{\ln(0.8)}\approx 7.213\). The minimum is \(8\) questions. 5. For part b), require \((0.8)^n\le 0.05\). 6. Thus, \(n\ge \frac{\ln(0.05)}{\ln(0.8)}\approx 13.425\). The minimum is \(14\) questions.

Answer

a) \(8\) questions b) \(14\) questions
Binomial thresholds and unknown parameters 17
An electronics manufacturer finds that each microchip independently has a \(5\%\) chance of being defective. The chips are shipped in packages of \(60\). a) Find the probability that more than \(10\%\) of the microchips in a randomly selected package are defective. Round to four decimal places. b) During quality control, the chips in a package are tested one at a time. Describe an event in context whose probability is represented by \(\sum_{k=0}^{4}(0.95)^k(0.05).\) c) What minimum proportion of the manufacturer's microchips must be nondefective so that a sample of \(30\) independent chips contains no defective chips with probability at least \(80\%\)? Give the answer as a percentage to three decimal places.

Hints

- In part a), convert the percentage threshold to the first qualifying whole-number defect count. - In part b), interpret one term \(q^kp\) before interpreting the sum. - In part c), express the probability that all \(30\) chips are nondefective.

Solution

1. For part a), let \(X\sim\operatorname{Bin}(60,0.05)\). More than \(10\%\) of \(60\) means at least \(7\) defective chips. 2. \(P(X\ge7)\approx0.0297.\) 3. In part b), \((0.95)^k(0.05)\) is the probability that the first defective chip occurs on test \(k+1\). Summing from \(k=0\) through \(4\) gives the probability that the first defective chip is found within the first \(5\) tests. 4. For part c), let \(q\) be the nondefective proportion. The requirement is \(q^{30}\ge0.80.\) Hence, \(q\ge(0.80)^{1/30}\approx0.992589,\) or approximately \(99.259\%\).

Answer

a) \(0.0297\). b) The first defective chip is found during one of the first \(5\) tests. c) Approximately \(99.259\%\) must be nondefective.
Binomial thresholds and unknown parameters 18
In an online game, each treasure chest independently has a \(12\%\) probability of containing a rare item. A player opens \(40\) chests. a) Find the probability that the player receives at most \(3\) rare items. Round to four decimal places. b) Describe an event in context whose probability is \(1-(0.88)^{15}\). c) What is the minimum number of treasure chests a player must open so that the probability of receiving at least \(1\) rare item is at least \(99\%\)?

Hints

- Part a is a cumulative binomial probability. - In part b), interpret \(0.88\) as the no-rare-item probability. - For part c), use the complement of receiving no rare items.

Solution

1. For part a), let \(X\sim\operatorname{Bin}(40,0.12)\). 2. \(P(X\le3)=\sum_{k=0}^{3}\binom{40}{k}(0.12)^k(0.88)^{40-k}\approx0.2768.\) 3. For part b), \((0.88)^{15}\) is the probability of receiving no rare items in \(15\) chests. Its complement is the probability of receiving at least one rare item. 4. For part c), require \(1-(0.88)^n\ge0.99.\) Thus, \(n\ge\frac{\ln(0.01)}{\ln(0.88)}\approx36.03,\) so the minimum integer is \(37\).

Answer

a) \(0.2768\). b) At least one rare item is found in \(15\) chests. c) \(37\) chests.
Binomial thresholds and unknown parameters 19
The random variable \(X\) has a binomial distribution with \(n=100\) and success probability \(p\). Find all values of \(p\) with \(0<p<1\) for which \(P(X=45)<P(X=55)\).

Hints

- Write the binomial formula for each probability. - Compare the two binomial coefficients. - Identify the positive factors common to both sides of the inequality. - Remember that both bases of the powers are positive.

Solution

1. Write the two binomial probabilities and compare them: \(\binom{100}{45}p^{45}(1-p)^{55}<\binom{100}{55}p^{55}(1-p)^{45}\). 2. Since \(\binom{100}{45}=\binom{100}{55}\) and all factors are positive for \(0<p<1\), divide by the common positive factor \(\binom{100}{45}p^{45}(1-p)^{45}\). This gives \((1-p)^{10}<p^{10}\). 3. Both bases are positive, and raising positive numbers to the tenth power preserves their order. Thus, \(1-p<p\). 4. Solving gives \(1<2p\), so \(p>0.5\). Combining this with \(p<1\) gives \(0.5<p<1\).

Answer

\(0.5<p<1\), or \(p\in(0.5, 1)\).
Binomial thresholds and unknown parameters 20
A random variable \(Y\) has a binomial distribution with \(n=3\) and \(p\in(0, 1)\). Find all values of \(p\) for which the probability of no successes is at least as great as the probability of exactly one success.

Hints

- Translate “at least as great as” into an inequality. - Substitute \(n=3\) and the relevant values of \(k\) into the binomial probability formula. - Simplify by dividing by a positive power of \(1-p\). - Intersect your solution with the given domain for \(p\).

Solution

1. Translate the condition as \(P(Y=0)\ge P(Y=1)\). 2. Substitute the binomial probabilities: \(\binom{3}{0}p^0(1-p)^3\ge\binom{3}{1}p(1-p)^2\). 3. Since \(p<1\), the factor \((1-p)^2\) is positive. Dividing by it gives \(1-p\ge3p\). 4. Solving the inequality gives \(1\ge4p\), so \(p\le0.25\). 5. Combining this result with \(p\in(0, 1)\) gives \(0<p\le0.25\).

Answer

\(0<p\le0.25\), or \(p\in(0, 0.25]\).
Binomial thresholds and unknown parameters 21
A binomial random variable \(X\) has parameters \(n=80\) and \(p=0.3\). Find the narrowest interval of the form \([\mu-k,\mu+k]\), symmetric about the mean, that contains at least \(95\%\) of the probability. Assume \(k\) is a nonnegative integer.

Hints

- Find the mean, which is the center of the interval. - Express an interval probability as a difference of cumulative probabilities. - Test integer distances from the mean systematically. - Use unrounded cumulative values when subtracting.

Solution

1. The mean is \(\mu=np=80\cdot 0.3=24\). 2. Seek the smallest nonnegative integer \(k\) such that \(P(24-k\le X\le 24+k)\ge 0.95\). 3. For \(k=7\), \(P(17\le X\le 31)\approx 0.9337<0.95\). 4. For \(k=8\), \(P(16\le X\le 32)\approx 0.9628\ge 0.95\). 5. Therefore, \(k=8\), and the narrowest qualifying symmetric interval is \([16,32]\).

Answer

\([16,32]\)
Binomial thresholds and unknown parameters 22
A binomial random variable \(Y\) has parameters \(n=150\) and \(p=0.2\). Find the narrowest interval of the form \([\mu-c,\mu+c]\), symmetric about the mean, that contains at least \(90\%\) of the probability. Assume \(c\) is a nonnegative integer.

Hints

- Find the mean, which is the center of the interval. - A symmetric interval has endpoints the same distance from the mean. - Use a binomial cumulative distribution function to evaluate each interval probability. - When subtracting cumulative probabilities, make sure the lower endpoint remains included.

Solution

1. The mean is \(\mu=np=150\cdot 0.2=30\). 2. Seek the smallest nonnegative integer \(c\) such that \(P(30-c\le Y\le 30+c)\ge 0.90\). 3. For \(c=7\), \(P(23\le Y\le 37)\approx 0.8751<0.90\). 4. For \(c=8\), \(P(22\le Y\le 38)\approx 0.9182\ge 0.90\). 5. Therefore, \(c=8\), and the narrowest qualifying symmetric interval is \([22,38]\).

Answer

\([22,38]\)
Binomial thresholds and unknown parameters 23
An agricultural researcher studies the germination of a seed variety. In \(150\) trials, each using \(5\) seeds, the number \(k\) of seeds that germinated was recorded. <table> <tr><td>Number germinated \(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Frequency \(n_k\)</td><td>\(4\)</td><td>\(21\)</td><td>\(50\)</td><td>\(48\)</td><td>\(22\)</td><td>\(5\)</td></tr> </table> 1. Find the mean number \(\bar{k}\) of germinated seeds per trial. 2. Assume the number germinated follows \(\operatorname{Bin}(5,p)\). Estimate \(p\) by setting the observed mean equal to the binomial mean. 3. Under the fitted model, find the probability that exactly \(2\) seeds germinate, rounded to four decimal places. Compare it with the observed relative frequency, also to four decimal places.

Hints

- Use a weighted mean for the frequency table. - Relate the observed mean to \(np\). - Use the fitted \(p\) in the binomial point probability. - Compare like quantities: model probability and observed relative frequency.

Solution

1. \(\bar{k}=\frac{0(4)+1(21)+2(50)+3(48)+4(22)+5(5)}{150}=2.52\). 2. Since the binomial mean is \(5p\), set \(5p=2.52\), giving \(p=0.504\). 3. \(P(X=2)=\binom{5}{2}(0.504)^2(0.496)^3\approx0.3100\). 4. The observed relative frequency is \(\frac{50}{150}=\frac{1}{3}\approx0.3333\), about \(0.0234\) higher than the model probability.

Answer

1. \(\bar{k}=2.52\). 2. \(p=0.504\). 3. Model: \(0.3100\); observed: \(0.3333\). The observed frequency is higher by about \(0.0234\).
Binomial thresholds and unknown parameters 24
An electronics manufacturer tests components in groups of \(3\). For \(200\) groups, the number \(k\) of defective components per group was recorded. <table> <tr><td>Number defective \(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>Number of groups \(n_k\)</td><td>\(130\)</td><td>\(55\)</td><td>\(12\)</td><td>\(3\)</td></tr> </table> 1. Find the overall proportion of defective components in the sample. Use it as an estimate of the defect probability \(p\) for one component. 2. Assume the number of defects per group follows \(\operatorname{Bin}(3,p)\). Using the estimate from part 1, find the probability that a group contains at least \(1\) defective component. Round to four decimal places. 3. Compare the model probability with the observed relative frequency of groups containing at least \(1\) defect, reporting both to four decimal places.

Hints

- Convert the grouped counts into a total number of defective components first. - Use the complement of no defects in a group. - Compare the fitted-model probability with the observed fraction of affected groups.

Solution

1. The total number of defective components is \(0(130)+1(55)+2(12)+3(3)=88\). Out of \(600\) tested components, \(p=\frac{88}{600}=\frac{11}{75}\approx0.1467\). 2. \(P(X\ge1)=1-\left(\frac{64}{75}\right)^3\approx0.3786\). 3. The observed relative frequency is \(\frac{55+12+3}{200}=0.3500\). The fitted model is about \(0.0286\) higher.

Answer

1. \(p=\frac{11}{75}\approx0.1467\). 2. \(0.3786\). 3. Model: \(0.3786\); observed: \(0.3500\). The model predicts a proportion about \(0.0286\) higher.
Binomial thresholds and unknown parameters 25
A binomial random variable \(X\) has success probability \(p=0.5\) and expected value \(E(X)=6\). a) Determine the number of trials \(n\). b) Calculate \(P(X=4)\) and \(P(X=8)\), rounded to four decimal places. c) Use the binomial probability formula \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) to explain why the histogram of any binomial distribution with \(p=0.5\) is symmetric about \(k=\frac{n}{2}\).

Hints

- Relate the expected value to the number of trials and the success probability. - Recall the symmetry identity for binomial coefficients. - Simplify the factors \(p^k\) and \((1-p)^{n-k}\) when \(p=0.5\). - Identify the midpoint between \(0\) and \(n\).

Solution

1. Since \(E(X)=np\), \(6=n\cdot0.5\), so \(n=12\). 2. \(P(X=4)=\binom{12}{4}(0.5)^4(0.5)^8=\binom{12}{4}(0.5)^{12}=\frac{495}{4096}\approx0.1208\). Because \(\binom{12}{8}=\binom{12}{4}\), \(P(X=8)\approx0.1208\) as well. 3. In general, when \(p=0.5\), \(P(X=k)=\binom{n}{k}(0.5)^n\). Since \(\binom{n}{k}=\binom{n}{n-k}\), it follows that \(P(X=k)=P(X=n-k)\). The values \(k\) and \(n-k\) are equally far from \(\frac{n}{2}\), establishing the symmetry.

Answer

a) \(n=12\) b) \(P(X=4)\approx0.1208\) and \(P(X=8)\approx0.1208\) c) For \(p=0.5\), \(P(X=k)=\binom{n}{k}(0.5)^n\). Because \(\binom{n}{k}=\binom{n}{n-k}\), \(P(X=k)=P(X=n-k)\), so the histogram is symmetric about \(k=\frac{n}{2}\).
Binomial thresholds and unknown parameters 26
A manufacturer reports that each electronic component independently has a \(4\%\) chance of having a particular manufacturing defect. A technician selects a random sample of \(n\) components. What is the minimum sample size needed so that the probability of finding at least \(1\) component with this defect is at least \(95\%\)?

Hints

- What is the complement of “at least \(1\) component is defective”? - How can you express the probability of exactly \(0\) defects in \(n\) trials? - Use logarithms when the unknown is in an exponent. - The sample size must be a whole number.

Solution

1. The probability of finding no defective components in a sample of size \(n\) is \((0.96)^n\). 2. Require \(1-(0.96)^n\ge 0.95\), which is equivalent to \((0.96)^n\le 0.05\). 3. Taking logarithms gives \(n\ge \frac{\ln(0.05)}{\ln(0.96)}\approx 73.385\). 4. Because \(n\) must be a whole number, the minimum sample size is \(74\).

Answer

At least \(74\) components must be selected.
Binomial thresholds and unknown parameters 27
A computer-chip manufacturer finds that \(8\%\) of its chips are defective. Let \(X_n\) be the number of defective chips in a sample of size \(n\), so \(X_n\sim\operatorname{Bin}(n,0.08)\). a) A sample of \(20\) chips is inspected. Find each probability, rounded to four decimal places. \(E_1\): None of the chips are defective. \(E_2\): At most \(2\) chips are defective. \(E_3\): Exactly \(17\) or \(18\) chips are nondefective. b) Find the minimum value of \(n\) needed so that the probability of finding at least \(1\) defective chip is at least \(99\%\).

Hints

- Translate nondefective counts into defective counts for \(E_3\). - “At most two” is a cumulative probability. - Use the complement of no defects for the inverse sample-size condition.

Solution

1. \(P(E_1)=P(X_{20}=0)=(0.92)^{20}\approx0.1887.\) 2. \(P(E_2)=P(X_{20}\le2)\approx0.7879.\) 3. Exactly \(17\) or \(18\) nondefective chips means exactly \(3\) or \(2\) defective chips, so \(P(E_3)=P(X_{20}=2)+P(X_{20}=3)\approx0.4125.\) 4. For part b), require \(1-(0.92)^n\ge0.99.\) Hence, \(n\ge\frac{\ln(0.01)}{\ln(0.92)}\approx55.230,\) so the minimum integer is \(56\).

Answer

a) \(P(E_1)\approx0.1887\), \(P(E_2)\approx0.7879\), \(P(E_3)\approx0.4125\). b) \(56\).
Binomial thresholds and unknown parameters 28
A gardener buys \(20\) seeds of a rare plant. The seeds germinate independently, but the germination probability \(p\) for each seed under the storage conditions is unknown. Find the minimum threshold for \(p\) that makes the probability that at least one seed germinates at least \(95\%\). Give an exact expression and a decimal approximation to six decimal places.

Hints

- Use the complement of “at least one success.” - Write the probability of exactly zero successes. - Rearrange the inequality so that the unknown probability is isolated. - Use a twentieth root to undo the twentieth power.

Solution

1. Let \(X\sim\operatorname{Bin}(20,p)\) be the number of seeds that germinate. 2. Use the complement: \(P(X\ge1)=1-P(X=0)=1-(1-p)^{20}\). 3. The requirement \(1-(1-p)^{20}\ge0.95\) is equivalent to \((1-p)^{20}\le0.05\). 4. Taking the twentieth root gives \(1-p\le\sqrt[20]{0.05}\), so \(p\ge1-\sqrt[20]{0.05}\). 5. The threshold is \(1-\sqrt[20]{0.05}\approx0.139108\), or approximately \(13.9108\%\).

Answer

\(p\ge1-\sqrt[20]{0.05}\). The threshold is approximately \(0.139108\), or \(13.9108\%\).
Binomial thresholds and unknown parameters 29
A new neighborhood has \(150\) households that use the same internet provider. At a randomly selected time during the evening, each household independently uses the service at full bandwidth with probability \(0.20\). Let \(X\) be the number using full bandwidth at that time. What is the minimum number of households the network must be able to serve at full bandwidth simultaneously so that demand is met with probability at least \(95\%\)?

Hints

- Define the random variable and identify its possible values. - Translate the capacity requirement into a cumulative-probability inequality. - Use the mean and spread to identify reasonable values to test. - Check consecutive capacity values to verify the minimum.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(150,0.20)\). 2. Seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.95\). 3. The mean and standard deviation are \(\mu=np=30\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{24}\approx 4.899\), which help locate the threshold. 4. Using cumulative binomial probabilities, \(P(X\le 37)\approx 0.9341<0.95\), while \(P(X\le 38)\approx 0.9554\ge 0.95\). 5. Therefore, the network must support at least \(38\) households simultaneously at full bandwidth.

Answer

The network must support at least \(38\) households simultaneously.
Binomial thresholds and unknown parameters 30
At a coworking space, \(15\) members are present. Each member uses a soundproof phone booth for exactly \(15\) of the \(60\) minutes in a representative hour. At a randomly selected time in that hour, assume the \(15\) events “member \(i\) is using a booth” are independent. Thus each member is using a booth with probability \(15/60=0.25\). Let \(X\) be the number of booths in use. a) Find the probability that at most \(4\) booths are needed. Round to four decimal places. b) Find the probability that more than \(6\) booths are needed. Round to four decimal places. c) What is the minimum number of booths needed so that demand can be met with probability at least \(95\%\)?

Hints

- Convert one member's booth-use time into a probability at a randomly selected time. - Use a complement for “more than \(6\).” - For the capacity decision, compare consecutive cumulative probabilities around the target.

Solution

1. Each member is using a booth with probability \(p=15/60=0.25\), and the simultaneous-use indicators are assumed independent. Hence, \(X\sim\operatorname{Bin}(15,0.25)\). 2. \(P(X\le4)\approx0.6865\). 3. \(P(X>6)=1-P(X\le6)\approx0.0566\). 4. For the capacity requirement, seek the smallest integer \(k\) such that \(P(X\le k)\ge0.95\). 5. Since \(P(X\le6)\approx0.9434<0.95\) and \(P(X\le7)\approx0.9827\ge0.95\), the minimum is \(7\) booths.

Answer

a) \(0.6865\). b) \(0.0566\). c) \(7\) booths.
Binomial thresholds and unknown parameters 31
In a warehouse, \(20\) employees work at packing stations. Each employee uses a powered pallet jack for exactly \(6\) of the \(60\) minutes in a representative hour. At a randomly selected time in that hour, assume the \(20\) events “employee \(i\) is using a pallet jack” are independent. Thus each employee is using one with probability \(6/60=0.10\). Let \(X\) be the number of pallet jacks needed. a) Find the probability that exactly \(2\) pallet jacks are needed. Round to four decimal places. b) Find the probability that more than \(4\) pallet jacks are needed. Round to four decimal places. c) What is the minimum number of pallet jacks that must be available so that demand is met with probability at least \(99\%\)?

Hints

- Convert the six-minute usage interval into a probability at a randomly selected time. - Distinguish an exact-count event from an upper-tail event. - Use a complement for “more than \(4\).” - Verify the minimum capacity by comparing consecutive cumulative probabilities around \(0.99\).

Solution

1. Each employee is using a pallet jack with probability \(p=6/60=0.10\), and the simultaneous-use indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(20,0.10)\). 2. For part a), \(P(X=2)=\binom{20}{2}(0.10)^2(0.90)^{18}\approx0.2852\). 3. For part b), \(P(X>4)=1-P(X\le4)\approx0.0432\). 4. For part c), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.99\). 5. Since \(P(X\le5)\approx0.9887<0.99\) and \(P(X\le6)\approx0.9976\ge0.99\), the minimum is \(6\).

Answer

a) \(0.2852\). b) \(0.0432\). c) \(6\) pallet jacks.
Binomial thresholds and unknown parameters 32
An office has \(50\) employees who share licenses for specialized software. Each employee uses the software for exactly \(15\) of the \(60\) minutes in a representative hour. At a randomly selected time in that hour, assume the \(50\) events “employee \(i\) is using the software” are independent. Thus each employee is using it with probability \(15/60=0.25\). What is the minimum number of licenses needed so that the probability that at least one employee cannot access the software is at most \(2\%\)?

Hints

- Convert the exact 15-minute usage duration into a random-instant occupancy probability. - Translate a shortage into an inequality involving simultaneous users and licenses. - Compare consecutive cumulative probabilities to prove the minimum.

Solution

1. Each employee is using the software with probability \(p=15/60=0.25\), and the simultaneous-use indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(50,0.25)\). 2. If \(k\) licenses are available, a shortage occurs when \(X>k\). The requirement is \(P(X>k)\le0.02\), equivalently \(P(X\le k)\ge0.98\). 3. \(P(X\le18)\approx0.9713<0.98\), while \(P(X\le19)\approx0.9861\ge0.98\). 4. Therefore, the minimum number of licenses is \(19\).

Answer

\(19\) licenses.
Binomial thresholds and unknown parameters 33
A fitness studio offers a \(60\)-minute circuit class for \(40\) participants. Each participant uses a rowing machine for exactly \(10\) minutes during the class. At a randomly selected time, assume the \(40\) events “participant \(i\) wants a rowing machine” are independent. Thus each participant wants a machine with probability \(10/60=1/6\). Let \(X\) be the number who want to use a rowing machine. a) Find the probability that more than \(10\) participants want to use a rowing machine at the same time. Round to four decimal places. b) What is the minimum total number of rowing machines needed so that the probability that someone must wait is at most \(5\%\)?

Hints

- Convert the ten minutes of use into a random-instant occupancy probability. - Use a complement for the current ten-machine shortage probability. - For the minimum capacity, compare consecutive cumulative probabilities around the target.

Solution

1. Each participant wants a rowing machine with probability \(p=10/60=1/6\), and the simultaneous-demand indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(40,1/6)\). 2. \(P(X>10)=1-P(X\le10)\approx0.0584\). 3. For part b), seek the smallest integer \(k\) such that \(P(X>k)\le0.05\), equivalently \(P(X\le k)\ge0.95\). 4. Since \(P(X\le10)\approx0.9416<0.95\) and \(P(X\le11)\approx0.9739\ge0.95\), the minimum is \(11\) rowing machines.

Answer

a) \(0.0584\). b) \(11\) rowing machines.
Binomial thresholds and unknown parameters 34
A manufacturer produces screws, and each screw independently has probability \(0.04\) of being defective. A quality-control sample contains \(1000\) screws. Use a normal approximation to find a symmetric interval about the binomial mean that contains the number of defective screws with probability approximately \(0.95\). Give the continuous endpoints to two decimal places and the possible integer counts in the interval.

Hints

- Find the binomial mean and standard deviation. - Check the large-count condition before using the normal approximation. - Use the critical value for a central \(95\%\) normal interval. - Distinguish the continuous approximation from the attainable integer counts.

Solution

1. Let \(X\sim\operatorname{Bin}(1000,0.04)\). The mean is \(\mu=np=40\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{38.4}\approx6.197\). Since \(np=40\) and \(n(1-p)=960\), the normal approximation is appropriate. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). 4. The radius is \(1.96(6.197\ldots)\approx12.15\), so the continuous interval is approximately \([27.85,52.15]\). 5. The integer counts inside this interval are \(28\) through \(52\).

Answer

Continuous approximation: \([27.85,52.15]\). Integer counts: \(28\) through \(52\).
Binomial thresholds and unknown parameters 35
At a light-bulb factory, \(4\%\) of bulbs have a minor cosmetic defect. Assume defects occur independently. a) Find the probability that at least one bulb in a random box of \(25\) bulbs has a cosmetic defect. Round to four decimal places. b) What is the minimum number of bulbs that must be inspected so that the probability of finding at least one bulb with a cosmetic defect is greater than \(90\%\)?

Hints

- Use the complement of finding no defective bulbs. - In part b), the unknown sample size appears in an exponent. - Solve the exponential inequality with logarithms and verify the neighboring integers.

Solution

1. The probability that one bulb does not have the defect is \(0.96\). 2. \(P(\text{at least one defect in 25})=1-(0.96)^{25}\approx0.6396\). 3. For part b), require \(1-(0.96)^n>0.90\), equivalently \((0.96)^n<0.10\). 4. Taking logarithms gives \(n>\frac{\ln(0.10)}{\ln(0.96)}\approx56.406\). 5. Therefore, the minimum integer is \(57\). Indeed, \(n=56\) gives a probability below \(0.90\), while \(n=57\) gives a probability above \(0.90\).

Answer

a) \(0.6396\). b) \(57\) bulbs.
Binomial thresholds and unknown parameters 36
A manufacturer knows that each precision component independently has probability \(0.04\) of having a minor cosmetic defect. A shipment contains \(2500\) components. Use a normal approximation to find a central \(95\%\) interval about the binomial mean for the number of defective components. Give continuous endpoints to two decimal places and the possible integer counts.

Hints

- Identify the binomial model. - Find its mean and standard deviation. - Check the large-count condition. - Use the critical value for a central \(95\%\) normal interval.

Solution

1. Let \(X\sim\operatorname{Bin}(2500,0.04)\). The mean is \(\mu=np=100\). 2. The standard deviation is \(\sigma=\sqrt{2500(0.04)(0.96)}=\sqrt{96}\approx9.798\). The large-count condition is satisfied. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). Thus, \(100\pm1.96(9.798\ldots)\approx[80.80,119.20]\). 4. The integer counts inside the interval are \(81\) through \(119\).

Answer

Continuous approximation: \([80.80,119.20]\). Integer counts: \(81\) through \(119\).
Binomial thresholds and unknown parameters 37
A quality-control inspector checks electronic components, each of which independently has a \(12\%\) chance of being defective. Let \(X\) be the number of defective components in a sample of size \(n\). a) Find the minimum number of components the inspector must check so that the probability of finding at least one defective component is at least \(90\%\). b) Use systematic trial or a table to find the minimum value of \(n\) for which the probability of finding at most one defective component is less than \(10\%\).

Hints

- Write each condition as an inequality involving \(P(X)\). - For part a), use the complement of finding at least one defective component. - For part b), add the probabilities for \(X=0\) and \(X=1\), then test integer values of \(n\). - When solving the logarithmic inequality, remember that \(\ln(0.88)\) is negative.

Solution

1. The probability a component is not defective is \(0.88\). 2. For a), require \(1-(0.88)^n\ge0.90\), or \((0.88)^n\le0.10\). 3. Taking logarithms gives \(n\ge\frac{\ln(0.10)}{\ln(0.88)}\approx18.012\). The probability of at least one defect increases with \(n\), so the minimum integer is \(19\). 4. For b), \(P(X\le1)=(0.88)^n+n(0.12)(0.88)^{n-1}\). 5. Checking consecutive values gives \(P(X\le1)\approx0.1100\) when \(n=30\), and \(P(X\le1)\approx0.0994\) when \(n=31\). 6. Adding more independent components cannot increase the chance of finding at most one defect, because the defect count can only stay the same or rise. Thus the first value below \(0.10\) is \(n=31\), the minimum.

Answer

a) \(n=19\). b) \(n=31\).
Binomial thresholds and unknown parameters 38
A binomial random variable \(X\) has \(n=12\). Find the success probability \(p\) such that \(P(X\le 1)=0.15\). Round your answer to the nearest hundredth.

Hints

- Which exact values of \(X\) are included in \(X\le 1\)? - Use the binomial probability formula for each of those values. - The resulting equation is not convenient to solve algebraically, so use a numerical solver or a table. - Test values of \(p\) and observe how the cumulative probability changes.

Solution

1. The condition is \(P(X\le 1)=P(X=0)+P(X=1)=0.15\). 2. Substitute the binomial probabilities: \((1-p)^{12}+12p(1-p)^{11}=0.15\). 3. Solve this equation numerically for \(0\le p\le 1\). 4. The solution is \(p\approx 0.2546086\), which rounds to \(p\approx 0.25\).

Answer

\(p\approx 0.25\)
Binomial thresholds and unknown parameters 39
A binomial random variable \(X\) has \(n=20\). Find the success probability \(p\) if \(P(X\ge 15)=0.05\). Round your answer to the nearest hundredth.

Hints

- Rewrite the “at least” probability using a complement. - What value must \(P(X\le 14)\) have? - Use a binomial cumulative-probability function and vary \(p\) to locate the solution. - As \(p\) increases, the probability of a large number of successes also increases.

Solution

1. Use the complement: \(P(X\ge 15)=1-P(X\le 14)=0.05\). 2. Therefore, \(P(X\le 14)=0.95\). 3. Solve the binomial cumulative-probability equation numerically for \(p\). 4. The solution is \(p\approx 0.5444176\). This value gives \(P(X\ge 15)=0.05\). 5. Rounded to the nearest hundredth, \(p\approx 0.54\).

Answer

\(p\approx 0.54\)
Binomial thresholds and unknown parameters 40
A basketball player has a free-throw success probability of \(p\). Assume the outcomes of the free throws are independent. Find \(p\), rounded to the nearest hundredth, if the probability that the player makes at most \(1\) of \(4\) free throws is \(65\%\).

Hints

- Which values of \(X\) are included in “at most \(1\)”? - Write an equation by adding the probabilities of \(0\) and \(1\) success. - Use a numerical solver or a table because \(p\) appears in several powers.

Solution

1. Let \(X\) be the number of made free throws. Then \(X\) has a binomial distribution with \(n=4\) and success probability \(p\). 2. The condition is \(P(X\le 1)=0.65\). 3. Substitute the binomial probabilities: \((1-p)^4+4p(1-p)^3=0.65\). 4. Solve the equation numerically on \(0\le p\le 1\). 5. The solution is \(p\approx 0.3009635\), so \(p\approx 0.30\) to the nearest hundredth.

Answer

The success probability must be approximately \(0.30\), or \(30\%\).
Binomial thresholds and unknown parameters 41
An athlete wants the probability of making at least \(13\) successes in \(15\) independent attempts to be at least \(75\%\). What is the minimum success probability \(p\) for each attempt? Round to the nearest thousandth.

Hints

- Identify the probability distribution for the number of successes. - Which values of \(X\) are included in “at least \(13\)”? - How does \(P(X\ge 13)\) change as \(p\) increases? - Use a graphing calculator, table, or numerical solver to find the boundary value of \(p\).

Solution

1. Let \(X\) be the number of successes. Then \(X\) has a binomial distribution with \(n=15\) and success probability \(p\). 2. The condition is \(P(X\ge 13)\ge 0.75\). 3. At the boundary, solve \(P(X\ge 13)=\sum_{k=13}^{15}\binom{15}{k}p^k(1-p)^{15-k}=0.75\). 4. A numerical solution gives \(p\approx 0.8836624\). 5. Rounded to the nearest thousandth, the minimum required success probability is \(p\approx 0.884\).

Answer

The success probability must be at least approximately \(0.884\).
Binomial thresholds and unknown parameters 42
A quality-control inspector checks samples of \(20\) LED bulbs. Each bulb independently has probability \(p\) of being defective. A sample is accepted only when at most \(1\) bulb is defective. Find the boundary value of \(p\) for which a sample is accepted with probability \(90\%\). Round the boundary value to the nearest thousandth.

Hints

- Translate “at most \(1\) defective bulb” into possible values of \(X\). - Write an equation for the cumulative binomial probability. - Decide whether decreasing \(p\) raises or lowers the acceptance probability. - Use a binomial probability function and a numerical solver to locate the boundary value.

Solution

1. Let \(X\) be the number of defective bulbs in a sample. Then \(X\) has a binomial distribution with \(n=20\) and defect probability \(p\). 2. At the boundary, \(P(X\le 1)=0.90\). 3. This gives \((1-p)^{20}+20p(1-p)^{19}=0.90\). 4. Solving numerically gives \(p\approx 0.0269141\). 5. Therefore, the boundary value is \(p\approx 0.027\) to the nearest thousandth. The acceptance probability is at least \(90\%\) for \(p\le 0.0269141\).

Answer

The boundary value is \(p\approx 0.027\). More precisely, the acceptance probability is at least \(90\%\) when \(p\le 0.0269141\).
Binomial thresholds and unknown parameters 43
The germination probability for each seed of a certain type of flower is \(25\%\), independently of the other seeds. What is the minimum number of seeds a gardener must plant so that the probability that at least \(3\) seeds germinate is at least \(90\%\)?

Hints

- Identify the known and unknown binomial parameters. - Use the complement of “at least \(3\)” to reduce the number of cases. - Evaluate the cumulative probability \(P(X\le2)\). - Test consecutive integer values of \(n\) to locate the minimum.

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with success probability \(p=0.25\) and an unknown number of trials \(n\). 2. Require \(P(X\ge3)\ge0.90\), equivalently \(P(X\le2)\le0.10\). 3. Evaluate \(P(X\le2)=\sum_{k=0}^{2}\binom{n}{k}(0.25)^k(0.75)^{n-k}\) for successive integer values of \(n\). 4. For \(n=19\), \(P(X\le2)\approx0.1113\), so \(P(X\ge3)\approx0.8887\). 5. For \(n=20\), \(P(X\le2)\approx0.0913\), so \(P(X\ge3)\approx0.9087\). 6. Adding a seed cannot decrease the number that germinate, so \(P(X\ge3)\) cannot decrease as \(n\) increases. Since \(19\) fails and \(20\) passes, \(20\) is the minimum.

Answer

The gardener must plant at least \(20\) seeds.
Binomial thresholds and unknown parameters 44
An archer has the same probability \(p\) of hitting the bull's-eye on each of \(10\) independent shots. The archer wants at least an \(80\%\) chance of hitting the bull's-eye on at least \(8\) shots. Find the threshold value \(p_0\) such that the goal is met when \(p\ge p_0\). Report \(p_0\) to three decimal places. Then, if \(p\) can be specified only in increments of \(0.01\), state the smallest value that meets the goal.

Hints

- Write the probability of at least \(8\) hits as the sum of the probabilities of exactly \(8\), \(9\), and \(10\) hits. - The unknown probability appears in several powers, so solve the threshold equation numerically. - After finding the threshold, test the adjacent hundredth values rather than rounding without checking the inequality.

Solution

1. Let \(X\sim\operatorname{Bin}(10,p)\). The condition is \(P(X\ge 8)\ge 0.80\). 2. The upper-tail probability is \(P(X\ge 8)=\binom{10}{8}p^8(1-p)^2+\binom{10}{9}p^9(1-p)+p^{10}\). 3. This probability increases as \(p\) increases. Solving \(P(X\ge 8)=0.80\) numerically gives \(p_0\approx 0.842365\), so the threshold rounded to three decimal places is \(0.842\). 4. At \(p=0.84\), \(P(X\ge 8)\approx 0.7936<0.80\), while at \(p=0.85\), \(P(X\ge 8)\approx 0.8202>0.80\). Therefore, the smallest hundredth that meets the goal is \(0.85\).

Answer

The threshold is \(p_0\approx 0.842\). If \(p\) is restricted to increments of \(0.01\), the smallest value that meets the goal is \(0.85\).
Binomial thresholds and unknown parameters 45
A shipment contains \(20\) electronic components. Each component independently meets quality standards with probability \(p\). The probability that at least \(18\) components meet the standards must be at least \(70\%\). Find the threshold value \(p_0\) such that the requirement is met when \(p\ge p_0\). Report \(p_0\) to three decimal places. Then, if \(p\) can be specified only in increments of \(0.01\), state the smallest value that meets the requirement.

Hints

- Write the probability of at least \(18\) acceptable components as the sum of the probabilities of exactly \(18\), \(19\), and \(20\). - Solve the threshold equation numerically because it cannot be isolated for \(p\) by elementary algebra. - After finding the threshold, test the adjacent hundredth values rather than rounding without checking the inequality.

Solution

1. Let \(X\sim\operatorname{Bin}(20,p)\). The condition is \(P(X\ge 18)\ge 0.70\). 2. The upper-tail probability is \(P(X\ge 18)=\binom{20}{18}p^{18}(1-p)^2+\binom{20}{19}p^{19}(1-p)+p^{20}\). 3. This probability increases as \(p\) increases. Solving \(P(X\ge 18)=0.70\) numerically gives \(p_0\approx 0.904057\), so the threshold rounded to three decimal places is \(0.904\). 4. At \(p=0.90\), \(P(X\ge 18)\approx 0.6769<0.70\), while at \(p=0.91\), \(P(X\ge 18)\approx 0.7334>0.70\). Therefore, the smallest hundredth that meets the requirement is \(0.91\).

Answer

The threshold is \(p_0\approx 0.904\). If \(p\) is restricted to increments of \(0.01\), the smallest value that meets the requirement is \(0.91\).
Binomial thresholds and unknown parameters 46
A quality-control inspector examines boxes containing \(50\) glass bottles each. Each bottle independently has probability \(p\) of containing a small air bubble. The probability that a box contains at most \(2\) bottles with air bubbles is exactly \(15\%\). Find \(p\). Round your answer to the nearest thousandth.

Hints

- Write the cumulative binomial probability \(P(X\le 2)\). - Substitute \(n=50\) and the possible values \(X=0,1,2\). - Use a numerical solver or a table because the equation is difficult to solve algebraically. - Test nearby values of \(p\) to locate where the cumulative probability is \(0.15\).

Solution

1. Let \(X\) be the number of bottles with air bubbles. Then \(X\) has a binomial distribution with \(n=50\) and probability \(p\). 2. The condition is \(P(X\le 2)=0.15\). 3. Therefore, \(\sum_{k=0}^{2}\binom{50}{k}p^k(1-p)^{50-k}=0.15\). 4. Solve the equation numerically. 5. The solution is \(p\approx 0.0919156\). For comparison, \(P(X\le 2)\approx 0.1550\) when \(p=0.091\) and \(P(X\le 2)\approx 0.1495\) when \(p=0.092\). 6. Rounded to the nearest thousandth, \(p\approx 0.092\).

Answer

The probability that a bottle contains a small air bubble is approximately \(0.092\), or \(9.2\%\).
Binomial thresholds and unknown parameters 47
Each computer chip produced by a manufacturer independently has a \(4\%\) chance of being defective. A sample of \(20\) chips is selected. a) What is the probability that exactly \(1\) chip is defective? Give an exact binomial expression and a decimal approximation to four decimal places. b) What is the probability that at most \(2\) chips are defective? Give a binomial-sum expression and a decimal approximation to four decimal places. c) What is the minimum number of chips that must be checked so that the probability of finding at least \(1\) defective chip is at least \(99\%\)?

Hints

- Use a binomial model for the fixed sample. - “At most \(2\)” includes \(0,1,2\). - For “at least \(1\),” use the complement of zero defects. - Use logarithms when the unknown sample size appears in an exponent.

Solution

1. For parts a) and b), \(X\sim\operatorname{Bin}(20,0.04)\). 2. \(P(X=1)=\binom{20}{1}(0.04)(0.96)^{19}\approx0.3683.\) 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{20}{k}(0.04)^k(0.96)^{20-k}\approx0.9561.\) 4. For part c), \(1-(0.96)^n\ge0.99\), so \(n\ge\frac{\ln(0.01)}{\ln(0.96)}\approx112.811.\) The minimum integer is \(113\).

Answer

a) \(\binom{20}{1}(0.04)(0.96)^{19}\approx0.3683\). b) \(\sum_{k=0}^{2}\binom{20}{k}(0.04)^k(0.96)^{20-k}\approx0.9561\). c) \(113\) chips.
Binomial thresholds and unknown parameters 48
A factory produces LED bulbs, and each bulb independently has a \(4\%\) probability of being defective. a) A package contains \(25\) bulbs. What is the probability that at least \(1\) bulb is defective? Give an exact expression and a decimal approximation to four decimal places. b) What is the minimum number of bulbs an inspector must test so that the probability of finding at least \(1\) defective bulb is at least \(90\%\)? c) What is the minimum number of bulbs needed so that the probability of finding at least \(2\) defective bulbs is at least \(90\%\)?

Hints

- Use complements for the “at least” events. - Use logarithms for part b). - For part c), test consecutive integer sample sizes around the threshold.

Solution

1. \(P(X\ge1)=1-(0.96)^{25}\approx0.6396\) for a sample of \(25\) bulbs. 2. For b), \(1-(0.96)^n\ge0.90\) gives \(n\ge56.4055\ldots\), so the smallest whole number is \(57\). 3. For c), require \(P(X\le1)=(0.96)^n+n(0.04)(0.96)^{n-1}\le0.10\). 4. For \(n=95\), the lower-tail probability is about \(0.10259\), which fails; for \(n=96\), it is about \(0.09931\), which meets the requirement. 5. Adding independent bulbs cannot decrease the number of defects found. Thus the probability of at least two defects is nondecreasing in \(n\), proving \(96\) is the minimum.

Answer

a) \(1-(0.96)^{25}\approx0.6396\). b) \(57\) bulbs. c) \(96\) bulbs.
Binomial thresholds and unknown parameters 49
During normal operation, \(5\%\) of a factory's components are defective. A random sample of \(20\) components is inspected, and defect outcomes are treated as independent. Let \(X\) be the number of defective components. a) Find the probability that the sample contains at least \(2\) defective components. Round to four decimal places. b) Find the smallest integer \(k\) for which the probability that the sample contains at least \(k\) defective components is less than \(1\%\).

Hints

- Use a complement for at least two defects. - For the threshold, test consecutive integer values of \(k\). - The first tail below \(1\%\) determines the answer.

Solution

1. \(X\sim\operatorname{Bin}(20,0.05)\). 2. \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642.\) 3. For the cutoff, compare consecutive upper tails. 4. \(P(X\ge4)\approx0.0159>0.01\), while \(P(X\ge5)\approx0.0026<0.01\). 5. Therefore, the smallest qualifying value is \(k=5\).

Answer

a) \(P(X\ge2)\approx0.2642\). b) \(k=5\).
Binomial thresholds and unknown parameters 50
For planning, a clinic assumes each patient independently has an \(85\%\) chance of responding to a treatment. A preliminary study must be large enough that the probability of a response in at least \(10\) patients is at least \(90\%\). Find the minimum number \(n\) of patients needed. Show how you used systematic testing or appropriate technology.

Hints

- Rewrite “at least \(10\)” using a complement so that a cumulative probability can be used. - Because \(n\) is unknown, begin with an integer somewhat larger than \(10\) and test consecutive values. - Consider how increasing the sample size affects the chance of observing at least \(10\) responses.

Solution

1. For a study of size \(n\), let \(X\sim\operatorname{Bin}(n,0.85)\). The requirement is \(P(X\ge 10)\ge 0.90\). 2. Using the complement, the condition is \(P(X\le 9)\le 0.10\). 3. For \(n=13\), \(P(X\le 9)\approx 0.1180>0.10\), so \(P(X\ge 10)\approx 0.8820<0.90\). 4. For \(n=14\), \(P(X\le 9)\approx 0.0467\le 0.10\), so \(P(X\ge 10)\approx 0.9533\ge 0.90\). 5. Adding a patient cannot reduce the chance of at least \(10\) responses, so smaller sample sizes cannot outperform \(n=13\). The minimum is \(n=14\).

Answer

The study needs at least \(14\) patients.
Binomial thresholds and unknown parameters 51
A certain flower seed germinates independently with probability \(0.80\). a) Twelve seeds are planted. Find the probabilities of these events, rounded to four decimal places: - \(C\): At most \(10\) seeds germinate. - \(D\): Exactly \(2\) seeds do not germinate. b) Explain the meaning of \(1-(0.8)^{12}\) in context. c) Among plants that germinate, each plant independently has a \(2\%\) probability of being affected by a particular pest. Assume these pest outcomes are independent across plants and independent of the germination outcomes. What is the minimum number of seeds that must be planted so that the probability of producing at least \(1\) affected plant is greater than \(95\%\)?

Hints

- For “at most \(10\),” a complement may shorten the calculation. - Exactly two failures means exactly ten germinations. - In part c), first find the one-seed probability of both germination and pest effect. - Use a complement and logarithms to solve the minimum-\(n\) condition.

Solution

1. For part a), let \(X\sim\operatorname{Bin}(12,0.80)\). 2. \(P(C)=P(X\le10)\approx0.7251.\) 3. Event \(D\) means exactly \(10\) seeds germinate, so \(P(D)=\binom{12}{10}(0.80)^{10}(0.20)^2\approx0.2835.\) 4. \((0.8)^{12}\) is the probability that all \(12\) seeds germinate, so \(1-(0.8)^{12}\) is the probability that at least one seed does not germinate. 5. For one planted seed, the probability of germinating and then producing an affected plant is \((0.80)(0.02)=0.016\). 6. For \(n\) seeds, require \(1-(0.984)^n>0.95.\) 7. Thus, \(n>\frac{\ln(0.05)}{\ln(0.984)}\approx185.731,\) so the minimum integer is \(186\).

Answer

a) \(P(C)\approx0.7251\) and \(P(D)\approx0.2835\). b) It is the probability that at least one of the twelve seeds does not germinate. c) \(186\) seeds.
Binomial thresholds and unknown parameters 52
A glass manufacturer finds that each drinking glass independently has an \(8\%\) chance of being defective. Three glasses are packed in each box. a) A box is acceptable if none of its \(3\) glasses is defective. Find the probability that a randomly selected box is acceptable. Give an exact expression and a decimal approximation to four decimal places. b) Assume boxes are independent. What is the minimum number of boxes that must be inspected so that the probability of finding at least \(1\) acceptable box is at least \(95\%\)? c) For a different packaging requirement, \(20\) glasses are placed in one box. What is the maximum individual-glass defect probability \(p\) for which the probability that the box contains no defective glasses is at least \(90\%\)? Give the exact inequality result and a percentage to three decimal places.

Hints

- Start with the probability that all glasses in one box are nondefective. - For part b), use the complement of finding no acceptable boxes. - For part c), express the all-nondefective probability in terms of \(1-p\) and solve the inequality.

Solution

1. A glass is nondefective with probability \(0.92\), so an acceptable three-glass box has probability \(p_B=(0.92)^3=0.778688\approx0.7787.\) 2. For \(n\) independent boxes, the probability of finding at least one acceptable box is \(1-(1-p_B)^n.\) Require \(1-(1-p_B)^n\ge0.95.\) This gives \(n\ge\frac{\ln(0.05)}{\ln(1-0.778688)}\approx1.986,\) so the minimum integer is \(2\). 3. For part c), require \((1-p)^{20}\ge0.90.\) Therefore, \(p\le1-(0.90)^{1/20}\approx0.005254.\) As a percentage, the maximum is approximately \(0.525\%\).

Answer

a) \((0.92)^3=0.778688\approx0.7787\). b) \(2\) boxes. c) \(p\le1-(0.90)^{1/20}\), so the maximum defect probability is approximately \(0.525\%\).
Binomial thresholds and unknown parameters 53
A seed company states that a certain variety of vegetable seed germinates independently with probability \(0.75\). a) A gardener plants a row of \(10\) seeds. Find the probability that the first \(3\) seeds germinate and at least \(4\) of the remaining \(7\) seeds germinate. Round to four decimal places. b) What is the minimum number of seeds the gardener must plant so that the probability that at least \(1\) seed germinates is at least \(99\%\)? c) Suppose the germination probability is unknown. What is the minimum value of \(p\) needed so that all \(10\) seeds in a packet germinate with probability at least \(95\%\)? Give the answer as a percentage to three decimal places.

Hints

- In part a), separate the fixed first-three outcome from the binomial count among the remaining seven. - For part b), use the complement event in which no seeds germinate. - For part c), write the all-ten-germinate probability as \(p^{10}\).

Solution

1. For part a), the probability that the first \(3\) seeds germinate is \((0.75)^3\). 2. If \(Y\sim\operatorname{Bin}(7,0.75)\), then \(P(Y\ge4)\approx0.9294.\) 3. By independence, the required probability is \((0.75)^3P(Y\ge4)\approx0.3921.\) 4. For part b), require \(1-(0.25)^n\ge0.99.\) Thus, \(n\ge\frac{\ln(0.01)}{\ln(0.25)}\approx3.322,\) so the minimum integer is \(4\). 5. For part c), require \(p^{10}\ge0.95.\) Therefore, \(p\ge(0.95)^{1/10}\approx0.994884.\) As a percentage, the minimum is approximately \(99.488\%\).

Answer

a) \(0.3921\). b) \(4\) seeds. c) Approximately \(99.488\%\).
Binomial thresholds and unknown parameters 54
A soldering defect occurs in \(5\%\) of electronic components. A random sample is selected for quality inspection, and defect outcomes are treated as independent. Find the minimum sample size \(n\) needed so that the probability of finding at least \(2\) components with soldering defects is at least \(90\%\).

Hints

- Model the number of defective components with a binomial random variable. - Use the complement event that at most one component has the defect. - Determine when the complement probability becomes small enough. - Test consecutive sample sizes with a calculator or statistical software.

Solution

1. For a sample of size \(n\), let \(X\sim\operatorname{Bin}(n,0.05)\). The requirement is \(P(X\ge2)\ge0.90\). 2. Using the complement, the condition becomes \(P(X\le1)\le0.10\). 3. Therefore, \((0.95)^n+n\cdot0.05\cdot(0.95)^{n-1}\le0.10\). 4. For \(n=76\), \(P(X\ge2)\approx0.8986<0.90\). For \(n=77\), \(P(X\ge2)\approx0.9027\ge0.90\). 5. Adding independent components cannot reduce the chance of finding at least two defects. The probability increases with \(n\), so \(76\) failing and \(77\) succeeding proves the minimum is \(77\).

Answer

The minimum sample size is \(77\) components.
Binomial thresholds and unknown parameters 55
A conservation project estimates that each released juvenile animal of a rare species has a \(15\%\) chance of surviving its first year in the wild. Assume the survival outcomes are independent. What is the minimum number of juvenile animals that must be released so that the probability that at least \(5\) survive the first year is at least \(99\%\)?

Hints

- Identify what counts as a success and its probability. - Express “at least \(5\)” as an inequality for a binomial random variable. - Use the complement event “at most \(4\)” to simplify the calculation. - Test consecutive values of the total number released to locate the minimum.

Solution

1. For \(n\) released animals, let \(X\sim\operatorname{Bin}(n,0.15)\). The requirement is \(P(X\ge5)\ge0.99\). 2. Using the complement, the condition becomes \(P(X\le4)\le0.01\). 3. For \(n=73\), \(P(X\le4)\approx0.0105>0.01\), so \(P(X\ge5)\approx0.9895<0.99\). 4. For \(n=74\), \(P(X\le4)\approx0.0094\le0.01\), so \(P(X\ge5)\approx0.9906\ge0.99\). 5. Including one more independently surviving animal cannot reduce the probability that at least five survive. Since \(73\) fails and \(74\) succeeds, the minimum is \(74\).

Answer

At least \(74\) juvenile animals must be released.
Binomial thresholds and unknown parameters 56
In a large shipment of electronic components, each component is defective independently with probability \(0.10\). A random sample of \(20\) components is selected. Let \(X\) be the number of defective components. a) Find the mean of \(X\). b) Find the shortest interval of consecutive integer values that contains the mean and has probability greater than \(90\%\).

Hints

- Use the mean formula for a binomial random variable. - Add point probabilities for consecutive integer counts near the mean. - Start with short intervals containing the mean and increase the number of included values only as needed. - Remember that a count-valued random variable takes only integer values.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(20,0.1)\), so its mean is \(\mu=np=20\cdot 0.1=2\). 2. The relevant point probabilities are \(P(X=0)\approx 0.1216\), \(P(X=1)\approx 0.2702\), \(P(X=2)\approx 0.2852\), \(P(X=3)\approx 0.1901\), \(P(X=4)\approx 0.0898\), and \(P(X=5)\approx 0.0319\). 3. The four-value intervals containing the mean are \([-1,2]\), \([0,3]\), \([1,4]\), and \([2,5]\). Their probabilities are approximately \(0.6770\), \(0.8670\), \(0.8352\), and \(0.5970\), respectively, so none exceeds \(0.90\). Therefore, no interval containing the mean with at most four integer values can exceed \(0.90\). 4. The five-value interval \([0,4]\) has probability \(P(0\le X\le 4)\approx 0.9568>0.90\). 5. Therefore, the shortest qualifying interval is \([0,4]\).

Answer

a) \(\mu=2\) b) \([0,4]\), with probability approximately \(0.9568\), or \(95.68\%\)
Binomial thresholds and unknown parameters 57
A company has \(20\) electric vehicles available through an employee car-sharing program. At a randomly selected time, each participating employee independently needs a vehicle with probability \(0.10\). What is the greatest number of employees who can participate so that the probability that every employee who needs a vehicle can get one is at least \(95\%\)?

Hints

- The success probability for needing a vehicle is given. - Find the largest \(n\) for which the cumulative probability through \(20\) remains at least \(0.95\). - Test nearby values of \(n\) using binomial technology. - Equivalently, the probability of needing more than \(20\) vehicles must remain at most \(5\%\).

Solution

1. If \(n\) employees participate, let \(X\sim\operatorname{Bin}(n,0.10)\) be the number who need a vehicle at the selected time. 2. The requirement is \(P(X\le 20)\ge 0.95\). 3. Evaluating nearby values gives \(P(X\le 20)\approx 0.952401\) for \(n=143\) and \(P(X\le 20)\approx 0.949330\) for \(n=144\). 4. For fixed success probability \(0.10\), \(P(X\le 20)\) decreases as \(n\) increases. Therefore, the greatest allowable number of employees is \(143\).

Answer

At most \(143\) employees can participate.

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