Free Math Worksheets

- Find the probability that one pollen grain lands in one specified cell. - Model the number of grains in that cell using repeated independent trials. - Multiply the one-cell probability by the number of cells to obtain the expected count.

1. For one selected cell, each pollen grain lands in that cell with probability \(p=\frac{1}{100}=0.01\). Let \(X\) be the number of pollen grains in the cell. Then \(X\) is binomial with \(n=120\) and \(p=0.01\). 2. \(P(X=0)=(0.99)^{120}\approx0.2994\). 3. \(P(X=1)=\binom{120}{1}(0.01)(0.99)^{119}\approx0.3629\). 4. \(P(X=2)=\binom{120}{2}(0.01)^2(0.99)^{118}\approx0.2181\). 5. By linearity of expectation, the expected number of cells containing exactly one pollen grain is \(100P(X=1)\approx36.29\).

a) \(P(X=0)\approx0.2994\) b) \(P(X=1)\approx0.3629\) c) \(P(X=2)\approx0.2181\) The expected number of cells containing exactly one pollen grain is about \(36.29\).

- List the \(25\) ordered pairs from two spins. - Group pairs that produce the same product. - Use the probability distribution to find the mean. - For independent spins, the expected product equals the product of the expected values. - In part c, find the new mean of one spin first.

1. There are \(5\cdot5=25\) equally likely ordered pairs. Counting pairs that produce each product gives the distribution in the answer. 2. From the distribution, \(E(X)=\frac{1\cdot1+2\cdot2+3\cdot2+4\cdot3+5\cdot2+6\cdot2+8\cdot2+9\cdot1+10\cdot2+12\cdot2+15\cdot2+16\cdot1+20\cdot2+25\cdot1}{25}=9\). 3. Alternatively, the spins are independent, so the expected product is the product of the single-spin means: \(E(X)=3\cdot3=9\). 4. After replacing \(5\) with \(k\), the mean of one spin is \(\frac{1+2+3+4+k}{5}=\frac{10+k}{5}\). 5. Require \(\left(\frac{10+k}{5}\right)^2=16\). Since the mean is positive, \(\frac{10+k}{5}=4\), giving \(k=10\).

a) <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(12\)</td><td>\(15\)</td><td>\(16\)</td><td>\(20\)</td><td>\(25\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{3}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td></tr> </table> b) \(E(X)=9\) c) \(k=10\)

- Model the number of landings on one specified section with a binomial distribution. - For more than \(2\) landings, use a complement. - For part b, first find the probability that one particular section is never reached. - Use indicator variables and linearity of expectation across all \(32\) sections.

1. Let \(X\) be the number of times the spinner lands on \(13\). Then \(X\) is binomial with \(n=100\) and \(p=\frac{1}{32}\). 2. \(P(X=0)=\left(\frac{31}{32}\right)^{100}\approx0.0418\). 3. \(P(X=1)=100\cdot\frac{1}{32}\left(\frac{31}{32}\right)^{99}\approx0.1348\). 4. \(P(X=2)=\binom{100}{2}\left(\frac{1}{32}\right)^2\left(\frac{31}{32}\right)^{98}\approx0.2153\). 5. \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.6081\). 6. For any particular section, the probability of being landed on at least once is \(1-\left(\frac{31}{32}\right)^{100}\). By linearity of expectation, the expected number of different sections reached is \(32\left[1-\left(\frac{31}{32}\right)^{100}\right]\approx30.66\).

a) \(P(X=0)\approx0.0418\), \(P(X=1)\approx0.1348\), \(P(X=2)\approx0.2153\), and \(P(X>2)\approx0.6081\) b) About \(30.66\) different sections

- For one error event, find the probability that it occurs on the specified server. - Model the number of errors on one server with a binomial distribution. - Use a complement for at least one error. - Apply linearity of expectation to the \(50\) server indicators.

1. For a specified server, the number of errors \(X\) is binomial with \(n=120\) and \(p=\frac{1}{50}=0.02\). 2. \(P(X=0)=0.98^{120}\approx0.0885\). 3. \(P(X=1)=120(0.02)(0.98)^{119}\approx0.2168\). 4. \(P(X=2)=\binom{120}{2}(0.02)^2(0.98)^{118}\approx0.2633\). 5. \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.4313\). 6. A specified server has probability \(1-0.98^{120}\) of experiencing at least one error. By linearity of expectation, the expected number of affected servers is \(50(1-0.98^{120})\approx45.57\).

a) \(P(X=0)\approx0.0885\), \(P(X=1)\approx0.2168\), \(P(X=2)\approx0.2633\), and \(P(X>2)\approx0.4313\) b) About \(45.57\) servers

- Treat each defect as one trial. - For one defect, find the probability that it lands on the specified chip. - Multiply the probability that one chip has exactly \(2\) defects by \(1000\). - Identify the independence and equal-probability assumptions.

1. For a specified chip, the number of defects \(X\) is binomial with \(n=50\) and \(p=\frac{1}{1000}=0.001\). 2. \(P(X=0)=(0.999)^{50}\approx0.9512\). 3. \(P(X=2)=\binom{50}{2}(0.001)^2(0.999)^{48}\approx0.0011676\). 4. By linearity of expectation, the expected number of chips with exactly \(2\) defects is \(1000\cdot0.0011676\approx1.17\). 5. The model assumes independent defect placements and the same probability for every chip. A success means that one defect lands on the specified chip, and the \(50\) defects are the \(50\) trials.

a) About \(0.9512\), or \(95.12\%\) b) About \(1.17\) chips c) Defect placements are independent and equally likely across chips. A success is one defect landing on the specified chip, and \(n=50\) is the number of defects.

- Use a complement for “at least \(3\).” - Think of the \(80\) colonies as separate trials. - Find the probability that one specified area is empty. - Use linearity of expectation across all \(200\) areas.

1. For one specified area, the number of colonies \(X\) is binomial with \(n=80\) and \(p=\frac{1}{200}=0.005\). 2. Use the complement: \(P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)\). 3. The needed probabilities are \(P(X=0)=(0.995)^{80}\approx0.6696\), \(P(X=1)=80(0.005)(0.995)^{79}\approx0.2692\), and \(P(X=2)=\binom{80}{2}(0.005)^2(0.995)^{78}\approx0.0534\). 4. Therefore, \(P(X\ge3)\approx0.0077\). 5. The expected number of empty areas is \(200P(X=0)\approx200(0.6696)=133.93\), or about \(134\). 6. The model treats each colony placement as an independent trial with the same probability of landing in the specified area.

a) About \(0.0077\), or \(0.77\%\) b) About \(134\) sample areas c) Each colony placement is modeled as independent, with a constant probability of landing in the specified area.

- For one specified number, find the probability that it does not occur on one roll. - Raise that probability to the number of rolls. - Use linearity of expectation for the number of missing values. - A logarithm can solve an equation with \(n\) in the exponent.

1. The probability of not rolling \(20\) on one roll is \(\frac{19}{20}=0.95\). Thus, the probability of no \(20\) in \(15\) rolls is \((0.95)^{15}\approx0.4633\). 2. For any specified number, the probability that it is still missing after \(n\) rolls is \((0.95)^n\). 3. By linearity of expectation, the expected number of missing values is \(20(0.95)^n\). Set this equal to \(4\): \(20(0.95)^n=4\). 4. Then \((0.95)^n=0.2\), so \(n=\frac{\ln(0.2)}{\ln(0.95)}\approx31.38\). The estimate is about \(31\) rolls.

a) \((0.95)^{15}\approx0.4633\), or about \(46.3\%\) b) About \(31\) rolls

- For one raisin, find the probability that it is placed in a specified roll. - Use a binomial model for the number of raisins in that roll. - Multiply the probability of more than \(4\) raisins by \(150\) for part b). - Use logarithms to solve the inequality in part c).

1. For a specified roll, each raisin has probability \(p=\frac{1}{150}\) of being placed in that roll. 2. With \(450\) raisins, \(X\) is binomial with \(n=450\) and \(p=\frac{1}{150}\). Thus, \(P(X=3)=\binom{450}{3}\left(\frac{1}{150}\right)^3\left(\frac{149}{150}\right)^{447}\approx0.2248\). 3. \(P(X>4)=1-\sum_{k=0}^{4}\binom{450}{k}\left(\frac{1}{150}\right)^k\left(\frac{149}{150}\right)^{450-k}\approx0.1842\). 4. The expected number of rolls with more than \(4\) raisins is \(150\cdot0.1842\approx27.63\), or about \(28\) rolls. 5. If \(n\) raisins are used, require \(P(X=0)=\left(\frac{149}{150}\right)^n\le0.05\). 6. Solving gives \(n\ge\frac{\ln(0.05)}{\ln(149/150)}\approx447.86\), so the minimum whole number is \(448\).

a) About \(0.2248\), or \(22.48\%\) b) About \(28\) rolls c) At least \(448\) raisins

- Convert the average number of defects per monitor into a total number of defects. - “At least one defect” is the complement of no defects. - In part b, use the observed proportion of defect-free monitors as an estimate of \(P(X=0)\).

1. An average of \(1.5\) defects on each of \(200\) monitors corresponds to \(n=300\) defects. For a specified monitor, each defect has probability \(\frac{1}{200}\) of landing there. 2. The probability of no defects is \(P(X=0)=\left(\frac{199}{200}\right)^{300}\approx0.2223\). 3. The probability of at least one defect is about \(0.7777\), so the expected number of affected monitors is \(200\cdot0.7777\approx155.54\), or about \(156\). 4. In part b, use the observed proportion \(\frac{90}{200}=0.45\) to estimate \(P(X=0)\). Solve \(\left(\frac{199}{200}\right)^n=0.45\). 5. This gives \(n=\frac{\ln(0.45)}{\ln(199/200)}\approx159.30\), so the estimated total is about \(159\) defects.

a) The probability of no pixel defects is about \(0.2223\), or \(22.23\%\). About \(156\) monitors are expected to have at least one defect. b) About \(159\) pixel defects

- For one specified value, the success probability on each roll is \(\frac{1}{6}\). - Use a binomial model for its number of appearances. - For part 3, use an indicator for whether each die value is missing. - Apply linearity of expectation to all six values.

1. For the specified number, the count \(X\) is binomial with \(n=6\) and \(p=\frac{1}{6}\). Thus, \(P(X=0)=\left(\frac{5}{6}\right)^6\approx0.3349\). 2. \(P(X=1)=\binom{6}{1}\frac{1}{6}\left(\frac{5}{6}\right)^5=\left(\frac{5}{6}\right)^5\approx0.4019\). 3. Each of the six values has probability \(\left(\frac{5}{6}\right)^6\) of being missing. By linearity of expectation, the expected number missing is \(6\left(\frac{5}{6}\right)^6\approx2.0094\). 4. One-third of \(6\) is \(2\), and \(2.0094>2\), so the expected number missing is slightly more than one-third of the possible values.

1. \(P(X=0)\approx0.3349\) 2. \(P(X=1)\approx0.4019\) 3. About \(2.01\) values are expected to be missing, which is slightly more than one-third of \(6\).

- Begin with one specified day and model the number of birthdays on that day. - Use linearity of expectation to extend a one-day probability to \(365\) days. - “More than one” is the complement of zero or exactly one. - The model assumes independent, uniformly distributed birthdays.

1. For one specified day, the number of employee birthdays \(X\) is binomial with \(n=500\) and \(p=\frac{1}{365}\). 2. \(P(X=0)=\left(\frac{364}{365}\right)^{500}\approx0.2537\). 3. By linearity of expectation, the expected number of days with no birthdays is \(365P(X=0)\approx92.59\). 4. The probability of exactly one birthday on a specified day is \(P(X=1)=500\cdot\frac{1}{365}\left(\frac{364}{365}\right)^{499}\approx0.3484\). 5. Therefore, \(P(X>1)=1-P(X=0)-P(X=1)\approx0.3979\). 6. The expected number of days with more than one birthday is \(365P(X>1)\approx145.23\).

a) About \(0.2537\), or \(25.37\%\) b) About \(92.59\) days c) About \(145.23\) days

- Focus first on one fixed integer. - Its number of appearances follows a binomial distribution. - Use indicator variables for whether each integer has the requested frequency. - Apply linearity of expectation across the \(100\) possible integers.

1. For one specified integer, its number of appearances \(X\) is binomial with \(n=100\) and \(p=0.01\). 2. The probability that it does not appear is \(P(X=0)=(0.99)^{100}\approx0.3660\). 3. By linearity of expectation, the expected number of missing integers is \(100(0.99)^{100}\approx36.60\). 4. The probability that one specified integer appears exactly twice is \(P(X=2)=\binom{100}{2}(0.01)^2(0.99)^{98}\approx0.1849\). 5. Therefore, the expected number of integers that appear exactly twice is \(100P(X=2)\approx18.49\).

a) About \(36.60\) values b) About \(18.49\) values

- Count the total equally likely outcomes for each experiment separately. - List all ordered pairs producing the requested sums. - “At least \(7\)” includes every sum of \(7\) or more. - Use linearity of expectation for each sum.

1. Experiment A has \(16\) equally likely ordered outcomes. Four produce a sum of \(5\), so \(P(X=5)=\frac{4}{16}=0.25\). 2. Experiment B has \(12\) equally likely ordered outcomes. Two produce a sum of \(5\), so \(P(Y=5)=\frac{2}{12}=\frac{1}{6}\approx0.1667\). Experiment A has the greater probability. 3. In Experiment A, sums of at least \(7\) occur in \(3\) outcomes, so \(P(X\ge7)=\frac{3}{16}=0.1875\). In Experiment B, they occur in \(3\) outcomes, so \(P(Y\ge7)=\frac{3}{12}=0.25\). The probabilities are not equal. 4. By linearity of expectation, \(E(X)=2.5+2.5=5\) and \(E(Y)=1.5+3.5=5\).

a) \(P(X=5)=0.25\) and \(P(Y=5)=\frac{1}{6}\approx0.1667\); Experiment A is greater. b) No. \(P(X\ge7)=0.1875\) and \(P(Y\ge7)=0.25\). c) \(E(X)=E(Y)=5\).

- For one selection, find the probability of choosing the specified song. - Use the complement of zero or one appearance for part b). - For part c, consider an indicator for whether each song is never selected. - Apply linearity of expectation across all \(50\) songs.

1. The number of times a specified song is played is binomial with \(n=50\) and \(p=\frac{1}{50}=0.02\). 2. \(P(X=0)=(0.98)^{50}\approx0.3642\). 3. \(P(X=1)=50\cdot0.02\cdot(0.98)^{49}=(0.98)^{49}\approx0.3716\). 4. Therefore, \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642\). 5. By linearity of expectation, the expected number of songs never played is \(50(0.98)^{50}\approx18.21\).

a) About \(0.3642\), or \(36.4\%\) b) About \(0.2642\), or \(26.4\%\) c) About \(18.21\) songs

- Determine the probability of each label on one spin. - List all possible sums from two spins. - Add the probabilities of all ordered pairs that produce the same sum. - Compute an expected value by multiplying each possible value by its probability. - You can also use the expected value of one spin and linearity of expectation.

1. On one spin, \(P(0)=\frac{3}{6}=\frac{1}{2}\), \(P(2)=\frac{2}{6}=\frac{1}{3}\), and \(P(5)=\frac{1}{6}\). 2. Combining the two independent spins gives \(P(X=0)=\frac{1}{4}\), \(P(X=2)=\frac{1}{3}\), \(P(X=4)=\frac{1}{9}\), \(P(X=5)=\frac{1}{6}\), \(P(X=7)=\frac{1}{9}\), and \(P(X=10)=\frac{1}{36}\). 3. The expected value is \(E(X)=0\left(\frac{1}{4}\right)+2\left(\frac{1}{3}\right)+4\left(\frac{1}{9}\right)+5\left(\frac{1}{6}\right)+7\left(\frac{1}{9}\right)+10\left(\frac{1}{36}\right)=3\). 4. Equivalently, one spin has expected value \(0\left(\frac{1}{2}\right)+2\left(\frac{1}{3}\right)+5\left(\frac{1}{6}\right)=1.5\), so two spins have expected sum \(2\cdot1.5=3\).

a) <table> <thead> <tr><th>\(k\)</th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th><th>\(5\)</th><th>\(7\)</th><th>\(10\)</th></tr> </thead> <tbody> <tr><th>\(P(X=k)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{36}\)</td></tr> </tbody> </table> b) \(E(X)=3\)