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Geometric distribution

Learning goals

Checks whether a setting satisfies geometric-model conditions.
Computes probabilities for the trial of first success.
Uses geometric mean or expected waiting time when appropriate.
Interprets results in context.

Difficulty: 1 – 5

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Geometric distribution 1

A library's self-checkout kiosk is restarted until it works. Each restart succeeds independently with the same probability \(p\), where \(0<p\le1\). Let \(X\) be the number of restarts needed for the first success. a) What is the smallest possible value of \(X\)? b) What does \(X=4\) mean in words? c) Is \(X=0\) possible?

Hints

- Focus on what \(X\) counts: the restart number of the first success. - What is the earliest restart that could succeed? - What must happen on each restart if the first success is on restart \(4\)?

Solution

1. A success can occur on the first restart, so the smallest possible value is \(1\). 2. The event \(X=4\) means the first three restarts fail and the fourth restart succeeds. 3. The value \(0\) is impossible because \(X\) counts the restart on which the first success occurs.

Answer

a) \(1\). b) The first three restarts fail and the fourth succeeds. c) No.

Geometric distribution 2

A fair die is rolled until a six appears. Let \(T\) be the roll number on which the first six appears. What does \(T=3\) mean? What is the smallest possible value of \(T\)?

Hints

- Focus on the first time the target result appears. - Consider what must have happened on all earlier rolls. - Could the target result occur immediately?

Solution

1. \(T=3\) means the first two rolls are not sixes and the third roll is a six. 2. The first roll could be a six, so the smallest value of \(T\) is \(1\).

Answer

\(T=3\) means the first six occurs on roll \(3\). The smallest possible value is \(1\).

Geometric distribution 3

A student flips a coin until it first lands on heads. Before that first head, there are three tails. Let \(T\) be the number of flips including the first head. What is \(T\), and why is it not \(3\)?

Hints

- Identify what counts as a failure and what counts as success. - The question asks for flips through the successful one, not failures alone. - Try listing the flips in order.

Solution

1. Three tails occur before the first head, so the outcome sequence ends \(TTTH\). 2. The first head is the fourth flip, which is included in the trial count. Thus, \(T=4\).

Answer

\(T=4\). The three failures are followed by one successful flip, which is also counted.

Geometric distribution 4

A test records the outcomes of three independent attempts, even if an earlier attempt succeeds. Write \(S\) for success and \(F\) for failure. Let \(T\) be the number of the first successful attempt. Which of the sequences \(FSF\), \(FFS\), and \(SSF\) describes \(T=3\)? Explain why the other two do not.

Hints

- Read each sequence from left to right. - Stop as soon as you reach the first success. - Do not count successes after the first one.

Solution

1. For \(T=3\), the first two attempts must fail and the third must succeed, giving \(FFS\). 2. \(FSF\) has its first success on attempt \(2\); \(SSF\) has its first success on attempt \(1\).

Answer

\(FFS\). In \(FSF\) and \(SSF\), the first success occurs before attempt \(3\).

Geometric distribution 5

A student spins the same fair four-section spinner repeatedly until the pointer first lands on a marked section. Spins are independent, and one of the four sections is marked. Let \(T\) be the spin number of the first marked result. Explain why \(T\) has a geometric distribution and state the success probability.

Hints

- What ends the sequence of spins? - Consider whether results on earlier spins alter the next spin. - Count how many equally likely spinner sections meet the stopping condition.

Solution

1. The target is a marked result, and each spin is a success or failure. 2. The spins are independent and each has the same success probability because the spinner does not change. 3. The chance of success is \(p=\frac{1}{4}\), and \(T\) counts trials until the first success, so it is geometric.

Answer

\(T\) is geometric because spins are independent, success probability stays constant, and it counts the first success. \(p=\frac{1}{4}\).

Geometric distribution 6

A coin is flipped independently, with the same chance of heads on each flip. Two students record different variables: - Rosa flips the coin exactly \(5\) times and counts the heads. - Omar keeps flipping until the first head and records which flip it occurs on. Which variable is binomial and which is geometric? Explain.

Hints

- Which student always completes the same number of flips? - Which student stops when a particular event first occurs? - Ask whether the variable counts successes or the number of the first successful trial.

Solution

1. Rosa counts successes in a fixed number of trials, so her variable is binomial. 2. Omar counts how many trials are needed to get the first success, so his variable is geometric.

Answer

Rosa's count is binomial; Omar's first-head flip number is geometric. The difference is a fixed number of trials versus waiting for the first success.

Geometric distribution 7

A fair coin is flipped until it first lands on heads. Let \(T\) be the number of the flip with the first head. The bar chart gives the probabilities for the first four values of \(T\). Read \(P(T=2)\) from the chart and explain what this event means.

Figure for problem 563472

Hints

- Find the bar directly above the requested trial number. - Think about what must happen before the first success. - The height of a probability bar gives the probability of that exact value.

Solution

1. The bar at \(T=2\) has height \(\frac{1}{4}\). 2. The event means the first flip was tails and the second was heads.

Answer

\(P(T=2)=\frac{1}{4}\). The first head occurs on the second flip.

Geometric distribution 8

A spinner has five equal sections, and exactly one section is blue. The spinner is spun independently until blue appears for the first time. Let \(X\) be the number of spins required. a) Find \(P(X=2)\) and \(P(X=3)\). b) Find the probability that at most three spins are required. c) Find the probability that more than four spins are required.

Hints

- For \(X=k\), think about the spins before the first blue result. - “At most three” means the first blue result occurs on one of the first three spins. - To need more than four spins, what must have happened during all four?

Solution

1. The success probability is \(p=\frac{1}{5}=0.20\), and the failure probability is \(0.80\). 2. \(P(X=2)=0.80\cdot0.20=0.16\). 3. \(P(X=3)=(0.80)^2(0.20)=0.128\). 4. \(P(X\le3)=1-P(\text{no blue in the first three spins})=1-(0.80)^3=0.488\). 5. More than four spins are required when the first four spins are not blue, so \(P(X>4)=(0.80)^4=0.4096\).

Answer

a) \(P(X=2)=0.16\); \(P(X=3)=0.128\). b) \(P(X\le3)=0.488\). c) \(P(X>4)=0.4096\).

Geometric distribution 9

A data-center security system has three detection stages in sequence. Each stage independently detects an unauthorized access attempt with probability \(0.80\). The attempt is stopped as soon as one stage detects it. Find the probability that an unauthorized attempt is 1. stopped by the first stage; 2. first stopped by the third stage; 3. detected by none of the three stages.

Hints

- What is the complement of a stage detecting the attempt? - What must happen at the first two stages for the third stage to be the first detector? - Which rule applies when independent events must occur in sequence?

Solution

1. Let \(p=0.80\) be the detection probability and \(q=1-p=0.20\) be the miss probability. 2. The first stage stops the attempt with probability \(p=0.80\). 3. For the third stage to be the first to detect the attempt, the first two stages must miss and the third must detect it. The probability is \(q^2p=(0.20)^2(0.80)=0.032\). 4. If no stage detects the attempt, all three stages miss. The probability is \(q^3=(0.20)^3=0.008\).

Answer

1. \(0.80\) 2. \(0.032\) 3. \(0.008\)

Geometric distribution 10

Inspections are independent. Use the tree to find the probability that the first detected defect occurs on the third inspection.

Figure for problem 537526

Hints

- Follow the path for the first detection on inspection \(3\). - Which branch must be taken at each earlier inspection? - Use the probabilities along the path to find the chance of the complete sequence.

Solution

1. The required sequence is no defect, no defect, defect. 2. From the tree, \(P(X=3)=(0.7)^2\cdot0.3=0.147\).

Answer

\(0.147=14.7\%\).

Geometric distribution 11

The number of auditions through the first callback is modeled by a geometric random variable \(X\) with mean \(5\). Louis says the standard deviation must also be \(5\) because geometric distributions are strongly right-skewed. Determine the success probability, find the actual standard deviation, and evaluate Louis's claim.

Hints

- Use the given long-run average wait to recover the trial success rate. - Treat center and spread as different properties of the distribution. - Check the claim numerically rather than relying on the shape alone.

Solution

1. Since \(E(X)=\frac{1}{p}=5\), the success probability is \(p=0.20\). 2. The standard deviation is \(\sigma_X=\frac{\sqrt{1-p}}{p}=\frac{\sqrt{0.80}}{0.20}=\sqrt{20}\approx4.47\). 3. Louis's claim is false; skewness does not force the mean and standard deviation to be equal.

Answer

\(p=0.20\) and \(\sigma_X=\sqrt{20}\approx4.47\). Louis's claim is false.

Geometric distribution 12

A device is tested repeatedly. On odd-numbered tests, the probability of passing is \(0.20\); on even-numbered tests, the probability of passing is \(0.50\). Test outcomes are independent. Let \(X\) be the test number of the first pass. Is \(X\) geometrically distributed? Explain, and find \(P(X=4)\).

Hints

- Compare the trial structure with all requirements of a geometric model. - List the outcome needed on each of the first four tests. - Use the stated independence after assigning the correct probability to each position.

Solution

1. A geometric model requires the same success probability on every trial. Here the probability alternates, so \(X\) is not geometric. 2. For \(X=4\), tests \(1\), \(2\), and \(3\) must fail and test \(4\) must pass. 3. The probability is \((0.80)(0.50)(0.80)(0.50)=0.16\).

Answer

No. The success probability is not constant. \(P(X=4)=0.16\).

Geometric distribution 13

A geometric process succeeds with probability \(0.27\) on each trial. To simulate one trial, Thandi chooses equally likely two-digit integers from \(00\) through \(99\). She labels \(00\) through \(27\) as success and all other integers as failure. a) Explain her error and give a correct assignment. b) Using the intended model, find the probability that more than \(3\) independent trials are needed. Round to four decimal places.

Hints

- When counting the integers in an interval, include both endpoints. - How many of the equally likely numbers should mean success? - What must happen in the first three trials for the wait to last longer?

Solution

1. The integers \(00\) through \(27\) include \(28\) equally likely outcomes, giving Thandi's assignment success probability \(0.28\). 2. One correct assignment is \(00\) through \(26\) for success and \(27\) through \(99\) for failure. 3. More than three trials means three consecutive failures, so the probability is \((0.73)^3=0.389017\approx0.3890\).

Answer

a) Thandi assigned \(28\) integers to success. Use \(00\)–\(26\) for success and \(27\)–\(99\) for failure. b) \(P(X>3)\approx0.3890\).

Geometric distribution 14

A quality check passes independently with probability \(0.45\) on each attempt. Let \(F\) be the number of failed checks before the first pass. Find the mean and standard deviation of \(F\). Give exact expressions and round decimals to two places.

Hints

- Does the count include the successful check? - How does counting only failures change the mean compared with counting all attempts? - Does subtracting the same number from every observation change its standard deviation?

Solution

1. Because \(F\) counts failures before the first success, its mean is \(E(F)=\frac{1-p}{p}=\frac{0.55}{0.45}=\frac{11}{9}\approx1.22\). 2. Subtracting one from the trials-through-success count does not change its standard deviation. 3. Thus, \(\sigma_F=\frac{\sqrt{1-p}}{p}=\frac{\sqrt{0.55}}{0.45}=\frac{2\sqrt{55}}{9}\approx1.65\).

Answer

Mean: \(\frac{11}{9}\approx1.22\) failures. Standard deviation: \(\frac{2\sqrt{55}}{9}\approx1.65\) failures.

Geometric distribution 15

A geometric random variable \(X\) counts attempts through the first success, with \(p=0.28\). Youssef writes \(P(X\ge5)=(0.72)^5\). Explain the error and find the correct probability.

Hints

- List the smallest waiting time included in the event. - Determine exactly which earlier trials must have failed. - Keep the difference between “at least” and “more than” clear.

Solution

1. The event \(X\ge5\) means that no success occurs on attempts \(1\) through \(4\). 2. It requires four failures, not five failures. 3. Therefore, \(P(X\ge5)=(0.72)^4\approx0.2687\).

Answer

Youssef counted one extra failure. The correct probability is \(P(X\ge5)=(0.72)^4\approx0.2687\).

Geometric distribution 16

A geometric random variable \(X\) counts trials through the first success, with \(p=0.05\). Its mean is \(20\) trials. Find the probability that the actual wait is longer than its mean, and interpret the result.

Hints

- Translate “longer than the mean” into a tail event using the integer-valued variable. - Determine how many initial failures that event requires. - Interpret the probability in light of the distribution's shape.

Solution

1. A wait longer than the mean is the event \(X>20\). 2. This requires the first \(20\) trials to fail, so \(P(X>20)=(0.95)^{20}\approx0.3585\). 3. Thus, even though the mean wait is \(20\), about \(35.85\%\) of waits exceed \(20\) trials because the distribution has a long right tail.

Answer

\(P(X>20)=(0.95)^{20}\approx0.3585\). About \(35.85\%\) of waits are longer than the mean.

Geometric distribution 17

A geometric random variable \(X\) counts trials through the first success and has mean \(8\). a) Find the success probability. b) Compare \(P(X=1)\) with \(P(X=8)\). Give exact values and decimal approximations to four decimal places. c) Explain why the most likely value can be much smaller than the mean.

Hints

- Use the given mean to work back to the chance of success on each trial. - For \(X=8\), how many failed trials occur before success? - How can the single most likely value differ from an average influenced by rare long waits?

Solution

1. Since \(E(X)=\frac{1}{p}=8\), the success probability is \(p=\frac{1}{8}=0.125\). 2. \(P(X=1)=\frac{1}{8}=0.1250\). 3. \(P(X=8)=\left(\frac{7}{8}\right)^7\left(\frac{1}{8}\right)=\frac{7^7}{8^8}\approx0.0491\). 4. Geometric point probabilities decrease as the attempt number increases, so the mode is \(1\). Occasional long waits pull the mean to the right.

Answer

a) \(p=\frac{1}{8}=0.125\). b) \(P(X=1)=\frac{1}{8}=0.1250\); \(P(X=8)=\frac{7^7}{8^8}\approx0.0491\). c) Long waits raise the mean even though attempt \(1\) is the most likely single waiting time.

Geometric distribution 18

A login attempt succeeds independently with probability \(0.80\). Let \(A\) be the number of successful logins among the next \(10\) attempts. Let \(B\) be the number of attempts required to get the first successful login. a) Which random variable is binomial, and what are its parameters? b) Which random variable is geometric, and what is its success probability? c) Explain the key difference between what \(A\) and \(B\) count.

Hints

- Ask whether the number of trials is fixed before the experiment begins. - Ask whether the random variable counts successes or counts waiting time. - Both models can use the same independent Bernoulli trials but summarize them differently.

Solution

1. \(A\) counts successes in a fixed number of \(10\) independent attempts, so \(A\sim\operatorname{Bin}(10,0.80).\) 2. \(B\) counts trials through the first success, so \(B\sim\operatorname{Geom}(0.80).\) 3. A binomial random variable counts how many successes occur in a fixed number of trials. A geometric random variable counts how long the process waits until the first success.

Answer

a) \(A\sim\operatorname{Bin}(10,0.80)\). b) \(B\sim\operatorname{Geom}(0.80)\). c) \(A\) counts successes in a fixed number of trials; \(B\) counts trials until the first success.

Geometric distribution 19

A machine restart succeeds independently with probability \(0.30\). The machine has already failed to start on five consecutive restarts. What is the probability that the next restart succeeds? Explain why the five previous failures do not change your answer.

Hints

- Re-read the independence condition. - Ask whether the outcome of an earlier restart changes the stated success probability on a later restart. - This is the memoryless feature of a geometric waiting-time process.

Solution

1. Each restart has the same success probability \(0.30\). 2. The restarts are independent, so the outcomes of earlier restarts do not change the probability of the next one. 3. Therefore, the probability that the next restart succeeds is \(0.30.\)

Answer

\(0.30\). Independence means the previous five failures do not change the success probability of the next restart.

Geometric distribution 20

A game is won on any one attempt with probability \(0.2\), independently of earlier attempts. Let \(T\) be the attempt on which the first win occurs. Find \(P(T=4)\).

Hints

- Translate “first win on the fourth attempt” into a sequence of failures and success. - Identify the chance of failure on one attempt. - Use independence to combine the probabilities in the required order.

Solution

1. For the first win to occur on attempt \(4\), the first three attempts must fail and the fourth must succeed. 2. The probability is \(P(T=4)=(0.8)^3(0.2)=0.1024\).

Answer

\(P(T=4)=0.1024\).

Geometric distribution 21

Each independent attempt to unlock a digital practice puzzle succeeds with probability \(0.2\). The player keeps trying until the first success. Let \(T\) be the number of the successful attempt. Find \(P(T>3)\) and explain what it means.

Hints

- A wait longer than three attempts tells you what happened on the first three. - Find the probability of failure on one attempt. - Independence tells you how to combine successive failures.

Solution

1. The event \(T>3\) means none of the first three attempts succeeds. 2. Each attempt fails with probability \(0.8\), so \(P(T>3)=(0.8)^3=0.512\).

Answer

\(P(T>3)=0.512\). The first three attempts all fail.

Geometric distribution 22

A fair coin is flipped independently until heads first appears. Let \(T\) be the number of the first head. Find the probability that the first head appears by the third flip by adding the probabilities for each possible first-head flip.

Hints

- List which trial numbers qualify for “by the third flip.” - Find the probability that the first success occurs on each one. - Add the probabilities of these mutually exclusive possibilities.

Solution

1. A first head on flip \(1\), \(2\), or \(3\) has probability \(\frac{1}{2}\), \(\frac{1}{4}\), or \(\frac{1}{8}\), respectively. 2. These events are mutually exclusive, so \(P(T\le3)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\).

Answer

\(P(T\le3)=\frac{7}{8}\).

Geometric distribution 23

A video game challenge is completed on any attempt with the same probability \(p\), independently of previous attempts. The average number of attempts up to and including the first completion is \(5\). Find \(p\).

Hints

- Relate the long-run average waiting time to the probability of success. - A larger success chance should make the wait shorter. - Work backward from the mean of the geometric distribution.

Solution

1. The mean number of trials until the first success in a geometric model is \(E(T)=\frac{1}{p}\). 2. Solve \(5=\frac{1}{p}\) to get \(p=\frac{1}{5}=0.20\).

Answer

\(p=0.20\).

Geometric distribution 24

A player has a \(0.25\) chance of winning each round of a game, independently of past rounds. The player has lost the first two rounds. A friend says, “Your chance of winning the next round must now be greater than \(0.25\).” Is the friend correct? Explain.

Hints

- Focus on the chance of the next round, not the chance of all three rounds together. - Consider what independence says about the past and the next event. - Ask whether the game's success probability has changed.

Solution

1. Independence means past losses do not change the chance of winning the next round. 2. The chance of winning round three remains \(0.25\), so the friend's claim is false.

Answer

No. The chance of winning the next round is still \(0.25\), because the rounds are independent.

Geometric distribution 25

A sticker booth makes a sale to each passerby with probability \(0.25\), independently of earlier passersby. The seller counts the people who do not buy anything before the first sale. What is the expected number of people who do not buy before the first sale?

Hints

- The usual geometric waiting count includes the successful attempt. - What is the average number of contacts through the first sale? - The question counts only unsuccessful contacts.

Solution

1. The number of people contacted through the first sale has geometric mean \(\frac{1}{0.25}=4\). 2. This count includes the buyer who makes the first purchase. 3. Subtract that one buyer to obtain an expected \(4-1=3\) nonbuyers before the first sale.

Answer

\(3\) people, on average.

Geometric distribution 26

A basketball player makes each free throw with probability \(0.20\), independently of earlier shots. The player has missed three times in a row. a) What is the probability of making the next shot? b) What is the probability of making at least one of the next two shots? c) Explain why the three earlier misses do not change either answer.

Hints

- Decide whether the player's success chance changes after a miss. - For two upcoming attempts, think about what it means not to succeed on either. - Keep the earlier shots separate from the two future shots.

Solution

1. Independence leaves the next-shot success probability at \(0.20\). 2. Missing both of the next two shots has probability \((0.80)^2=0.64\). 3. The probability of making at least one is \(1-0.64=0.36\). 4. Earlier results do not change the success probability of an independent future shot.

Answer

a) \(0.20\). b) \(0.36\). c) The shots are independent; previous misses do not alter the chances on future shots.

Geometric distribution 27

At a factory, each component independently has probability \(0.25\) of needing rework. Let \(X\) be the number of components inspected until the first component needing rework is found. Find the smallest number \(n\) of inspections for which the probability of finding at least one component needing rework is at least \(80\%\).

Hints

- Use the complement of finding at least one component needing rework. - Write an inequality involving the probability of no successes in \(n\) trials. - Test consecutive whole-number values near the threshold.

Solution

1. The probability that a component does not need rework is \(0.75\). 2. The probability of finding at least one component needing rework in \(n\) inspections is \(P(X\le n)=1-(0.75)^n\). 3. Require \(1-(0.75)^n\ge 0.80\), or \((0.75)^n\le 0.20\). 4. For \(n=5\), \(1-(0.75)^5\approx 0.7627\), which is less than \(0.80\). 5. For \(n=6\), \(1-(0.75)^6\approx 0.8220\), which meets the requirement. Therefore, the smallest value is \(n=6\).

Answer

\(n=6\)

Geometric distribution 28

An online retailer sends a promotional email to a new customer. The probability that the customer opens any one email is \(12\%\), independently of the other emails. The retailer sends one email per day until the customer opens one, but sends at most \(5\) emails total. Find the expected number of emails the customer receives. Round to two decimal places.

Hints

- What are the possible numbers of emails sent? - A fifth email is sent if the first four are not opened; does opening the fifth change how many are sent? - For fewer than five emails, consider when the first opening occurs. - How can you use the probabilities to find the average number sent?

Solution

1. Let \(X\) be the number of emails sent, so \(X\in\{1,2,3,4,5\}\). The chance an email is not opened is \(q=0.88\). 2. The probabilities are \(P(X=1)=0.12\), \(P(X=2)=0.88(0.12)=0.1056\), \(P(X=3)=0.88^2(0.12)=0.092928\), \(P(X=4)=0.88^3(0.12)=0.08177664\), and \(P(X=5)=0.88^4=0.59969536\). A fifth email is sent whenever the first four are not opened. 3. \(E(X)=1(0.12)+2(0.1056)+3(0.092928)+4(0.08177664)+5(0.59969536)=3.93556736\). 4. To two decimal places, \(E(X)\approx3.94\) emails.

Answer

About \(3.94\) emails.

Geometric distribution 29

At a carnival game, each independent spin has a \(25\%\) chance of winning. A player spins until winning once, but stops after at most \(4\) spins. Each spin costs \(\$1.50\). Find the expected total cost of one series of spins, rounded to the nearest cent.

Hints

- What are the possible numbers of spins in one game? - What must happen for the fourth spin to take place? - First find the expected number of spins, then connect spins to their cost.

Solution

1. Let \(X\) be the number of spins. Then \(X\in\{1,2,3,4\}\), with success probability \(p=0.25\) and failure probability \(q=0.75\). 2. The probabilities are \(P(X=1)=0.25\), \(P(X=2)=0.75(0.25)=0.1875\), \(P(X=3)=0.75^2(0.25)=0.140625\), and \(P(X=4)=0.75^3=0.421875\). 3. \(E(X)=1(0.25)+2(0.1875)+3(0.140625)+4(0.421875)=2.734375\) spins. 4. The expected cost is \(\$1.50\cdot2.734375=\$4.1015625\), which rounds to \(\$4.10\).

Answer

About \(\$4.10\).

Geometric distribution 30

A fair coin is flipped independently until the first head appears. Let \(T\) be the number of flips needed. a) List the possible values of \(T\). b) Find \(P(T=n)\) for any positive integer \(n\). c) Show that the probabilities for all possible values of \(T\) add to \(1\).

Hints

- What has to happen before the first head on flip \(n\)? - What is the probability of any one specified sequence of flips? - Compare the probabilities for success on the first, second, and third flips. How do they change?

Solution

1. The possible values are \(1,2,3,\dots\). The event \(T=n\) means \(n-1\) tails followed by one head. 2. Independence gives \(P(T=n)=\left(\frac{1}{2}\right)^{n-1}\cdot\frac{1}{2}=2^{-n}\). 3. The geometric series has sum \(\sum_{n=1}^{\infty}2^{-n}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\).

Answer

a) \(T\in\{1,2,3,\dots\}\). b) \(P(T=n)=2^{-n}\) for every positive integer \(n\). c) \(\sum_{n=1}^{\infty}2^{-n}=1\).

Geometric distribution 31

An emergency beacon transmits successfully on each attempt with probability \(0.65\). Attempts are independent. Find the smallest number of attempts needed so that the probability of at least one successful transmission is at least \(0.99\).

Hints

- Work with the complementary event that every attempt fails. - Express the target as an inequality involving the number of attempts. - After solving the inequality, check the neighboring integers.

Solution

1. The probability that all \(n\) attempts fail is \((1-0.65)^n=0.35^n\). 2. The requirement is \(1-0.35^n\ge0.99\), so \(0.35^n\le0.01\). 3. Taking logarithms gives \(n\ge\frac{\ln(0.01)}{\ln(0.35)}\approx4.387\). 4. The smallest integer is \(n=5\). Indeed, four attempts give \(1-0.35^4=0.98499375<0.99\), while five give \(1-0.35^5=0.9947478125\ge0.99\).

Answer

The beacon needs at least \(5\) attempts.

Geometric distribution 32

A box contains \(3\) green cards and \(5\) white cards. Cards are drawn one at a time without replacement until the first green card appears. Let \(X\) be the draw number of the first green card. Zahra claims that \(X\) has a geometric distribution with success probability \(\frac{3}{8}\). Is Zahra correct? Explain, and find \(P(X=3)\).

Hints

- Check what happens to the chance of success after an unsuccessful draw. - Compare the sampling process with the conditions required for repeated identical trials. - Describe the exact sequence of outcomes needed for the first success on draw \(3\).

Solution

1. The draws are not independent because cards are not replaced, and the probability of green changes after each white card. Therefore, \(X\) is not geometric. 2. For \(X=3\), the first two cards must be white and the third green. 3. The probability is \(\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{5}{28}\approx0.1786\).

Answer

No. The trials are not independent and the success probability is not constant. \(P(X=3)=\frac{5}{28}\approx0.1786\).

Geometric distribution 33

A network request succeeds independently on each attempt with probability \(0.18\). Let \(Y\) be the number of attempts through the first success, and let \(X\) be the number of failed attempts before the first success. a) State the relationship between \(X\) and \(Y\). b) Find \(P(X=4)\), rounded to four decimal places. c) Find the mean of each variable, rounded to two decimal places.

Hints

- Is the successful attempt included in each variable? - For four failures before the success, what sequence of outcomes is needed? - How does a one-unit shift affect the two expected values?

Solution

1. Because the successful attempt is included in \(Y\) but not in \(X\), \(Y=X+1\). 2. Four failures followed by success have probability \(P(X=4)=(0.82)^4(0.18)=0.0813819168\approx0.0814\). 3. The mean number of attempts through success is \(E(Y)=\frac{1}{0.18}\approx5.56\). 4. Thus, \(E(X)=E(Y)-1=\frac{0.82}{0.18}\approx4.56\).

Answer

a) \(Y=X+1\). b) \(P(X=4)\approx0.0814\). c) \(E(X)\approx4.56\) failures and \(E(Y)\approx5.56\) attempts.

Geometric distribution 34

For a geometric random variable \(X\) that counts trials through the first success, \(P(X>5)=0.20\). Find the success probability \(p\) for each trial and the mean of \(X\). Give exact expressions and round \(p\) to four decimal places and the mean to two decimal places.

Hints

- What does \(X>5\) tell you about the first five trials? - Relate the chance of five failures to the chance of failure on just one trial. - Once you know the success probability, how is the average waiting time found?

Solution

1. The event \(X>5\) means the first five trials are failures, so \((1-p)^5=0.20\). 2. Thus, \(1-p=0.20^{1/5}\), giving \(p=1-0.20^{1/5}\approx0.2752\). 3. The mean is \(E(X)=\frac{1}{p}=\frac{1}{1-0.20^{1/5}}\approx3.63\) trials.

Answer

\(p=1-0.20^{1/5}\approx0.2752\) and \(E(X)=\frac{1}{1-0.20^{1/5}}\approx3.63\) trials.

Geometric distribution 35

A rare signal is detected independently on each scan with probability \(0.12\). Eight scans have already failed to detect it. a) Given this information, find the probability that more than \(5\) additional scans are needed. Round to four decimal places. b) Find the expected number of additional scans through detection. Give an exact value and round to two decimal places.

Hints

- Does a run of failed scans change the success chance on the next independent scan? - Reframe the question as a new waiting time starting now. - Distinguish a probability that the wait exceeds a threshold from its mean length.

Solution

1. Independence means earlier failures do not change the success probability of the next scan. 2. More than five additional scans means five further failures, with probability \((0.88)^5=0.5277319168\approx0.5277\). 3. The expected number of additional scans through detection is \(\frac{1}{0.12}=\frac{25}{3}\approx8.33\).

Answer

a) \(0.5277\). b) \(\frac{25}{3}\approx8.33\) additional scans.

Geometric distribution 36

A geometric random variable \(X\) counts trials through the first success. It is known that \(\frac{P(X=5)}{P(X=4)}=0.65\). Find the success probability \(p\) and then find \(P(X=4)\), rounded to four decimal places.

Hints

- Write expressions for the two consecutive waiting-time probabilities. - What cancels when their ratio is simplified? - Once you know the trial rate, use the event of success on the fourth attempt.

Solution

1. For a geometric variable, \(P(X=5)=(1-p)^4p\) and \(P(X=4)=(1-p)^3p\). 2. Their ratio is \(1-p\), so \(1-p=0.65\) and \(p=0.35\). 3. Then \(P(X=4)=(0.65)^3(0.35)=0.09611875\approx0.0961\).

Answer

\(p=0.35\) and \(P(X=4)\approx0.0961\).

Geometric distribution 37

A geometric random variable \(X\) counts attempts through the first success, with success probability \(0.40\). Given that the first success occurs within the first \(4\) attempts, find the probability that it occurs on attempt \(2\). Give an exact fraction and a decimal rounded to four places.

Hints

- Translate the given condition into an event involving \(X\). - Is getting the first success on attempt \(2\) part of the event described by the condition? - Find the probability of the event and the probability of the condition separately.

Solution

1. The required conditional probability is \(P(X=2\mid X\le4)=\frac{P(X=2)}{P(X\le4)}\). 2. The numerator is \(P(X=2)=(0.60)(0.40)=0.24\). 3. The denominator is \(P(X\le4)=1-(0.60)^4=0.8704\). 4. Thus, \(P(X=2\mid X\le4)=\frac{0.24}{0.8704}=\frac{75}{272}\approx0.2757\).

Answer

\(P(X=2\mid X\le4)=\frac{75}{272}\approx0.2757\).

Geometric distribution 38

A diagnostic test gives a usable reading independently on each attempt with probability \(0.22\). Let \(X\) be the attempt number of the first usable reading. Find \(P(3\le X\le6)\).

Hints

- Translate the lower bound into a required string of early failures. - Remove outcomes in which the wait continues beyond the upper bound. - A difference of two tail probabilities can avoid adding several separate cases.

Solution

1. The event \(3\le X\le6\) means the first two attempts fail, but not all of the first six attempts fail. 2. The probability of at least two initial failures is \((0.78)^2\). 3. Subtract the probability of six initial failures: \(P(3\le X\le6)=(0.78)^2-(0.78)^6\approx0.3832\).

Answer

\(P(3\le X\le6)=(0.78)^2-(0.78)^6\approx0.3832\).

Geometric distribution 39

A geometric random variable \(X\) counts calls through the first answered call, with answer probability \(0.25\) on each independent call. Given that the first \(12\) calls were unanswered, find \(P(X\le15\mid X>12)\).

Hints

- Reinterpret the condition as a new starting point after a string of failures. - Count how many additional trials are allowed. - Use the complement of no success in those additional trials.

Solution

1. Given \(X>12\), the process restarts with the same success probability because of memorylessness. 2. The event \(X\le15\) then means at least one answer in the next three calls. 3. The probability is \(1-(0.75)^3=0.578125\approx0.5781\).

Answer

\(P(X\le15\mid X>12)=1-(0.75)^3\approx0.5781\).

Geometric distribution 40

A contestant repeats an independent challenge until the first success. Each attempt costs \(\$3.00\), the success probability is \(0.16\) per attempt, and a successful completion pays a single prize of \(\$40.00\). a) Find the expected total attempt cost. b) Find the expected net payoff. c) What prize would make the expected net payoff equal to \(\$0\)?

Hints

- First find the long-run average number of attempts through success. - Convert the expected attempt count into an expected total cost. - For break-even, compare the fixed prize with the expected cost.

Solution

1. The expected number of attempts is \(\frac{1}{0.16}=6.25\). 2. The expected total attempt cost is \(\$3.00(6.25)=\$18.75\). 3. The expected net payoff is \(\$40.00-\$18.75=\$21.25\). 4. A break-even prize must equal the expected cost, so it is \(\$18.75\).

Answer

a) \(\$18.75\). b) \(\$21.25\). c) \(\$18.75\).

Geometric distribution 41

A rescue beacon has the same independent probability \(p\) of connecting on each attempt. Let \(X\) be the number of attempts through the first connection. Find the minimum value of \(p\) that makes \(P(X\le8)\ge0.95\).

Hints

- Rephrase connection within eight attempts using its complement. - Translate the probability requirement into an inequality for repeated failures. - Keep the direction of the inequality in mind when solving for the success rate.

Solution

1. The probability of at least one connection in eight attempts is \(1-(1-p)^8\). 2. Require \(1-(1-p)^8\ge0.95\), so \((1-p)^8\le0.05\). 3. Thus, \(1-p\le0.05^{1/8}\), giving \(p\ge1-0.05^{1/8}\approx0.3123\).

Answer

The minimum success probability is \(p=1-0.05^{1/8}\approx0.3123\).

Geometric distribution 42

A technician can reset a router or run a connection repair. Attempts are independent within each approach, and each approach is repeated until it first succeeds. A reset succeeds with probability \(0.10\) per attempt; a repair succeeds with probability \(0.20\) per attempt. Let \(R\) count resets and \(C\) count repair attempts through the first success. a) For a reset and a repair, find the probability that the first success occurs on attempt \(5\). b) For each approach, find the probability that more than \(5\) attempts are needed. c) Explain why a repair can be more likely to succeed on exactly attempt \(5\) but less likely to need more than \(5\) attempts. Round numerical probabilities to four decimal places.

Hints

- What must happen before success on one specified attempt? - How is “more than five attempts” different from “success on the fifth”? - Think about how a larger chance of success changes the entire waiting-time distribution.

Solution

1. \(P(R=5)=(0.90)^4(0.10)=0.06561\approx0.0656\). 2. \(P(C=5)=(0.80)^4(0.20)=0.08192\approx0.0819\). 3. \(P(R>5)=(0.90)^5=0.59049\approx0.5905\). 4. \(P(C>5)=(0.80)^5=0.32768\approx0.3277\). 5. Repairs put more probability on early successes. That can raise the chance of success exactly on attempt \(5\) while reducing the chance of waiting longer.

Answer

a) Reset: \(0.0656\); repair: \(0.0819\). b) Reset: \(0.5905\); repair: \(0.3277\). c) A higher success rate shifts probability to earlier attempts. A single attempt and the entire tail beyond it need not change in the same direction.

Geometric distribution 43

A search algorithm originally succeeds independently with probability \(0.08\) per run. An update doubles the success probability to \(0.16\). Let \(X\) and \(Y\) count the runs through the first success before and after the update. Compare the means and standard deviations of \(X\) and \(Y\), rounding standard deviations to two decimal places. Does doubling \(p\) exactly halve both measures?

Hints

- Calculate center and spread separately for each success rate. - Compare how each formula depends on the success probability. - When one input doubles, do the other factors in the two formulas stay unchanged?

Solution

1. Before the update, \(E(X)=\frac{1}{0.08}=12.5\) and \(\sigma_X=\frac{\sqrt{0.92}}{0.08}\approx11.99\). 2. After the update, \(E(Y)=\frac{1}{0.16}=6.25\) and \(\sigma_Y=\frac{\sqrt{0.84}}{0.16}\approx5.73\). 3. Doubling \(p\) exactly halves the mean because the mean is \(\frac{1}{p}\). It does not exactly halve standard deviation because \(\sqrt{1-p}\) also changes.

Answer

Before: mean \(12.5\), standard deviation \(\approx11.99\). After: mean \(6.25\), standard deviation \(\approx5.73\). The mean is halved exactly, but the standard deviation is not.

Geometric distribution 44

A geometric random variable \(X\) counts trials through the first success, with \(p=0.20\). a) Find the median of \(X\), defined as the smallest integer \(m\) for which \(P(X\le m)\ge0.50\). b) Find \(P(X=m)\).

Hints

- Translate the median definition into a cumulative probability threshold. - Because the variable is discrete, check the first integer that crosses the threshold. - After finding the median value, treat the second part as an exact waiting-time event.

Solution

1. Solve \(1-(0.80)^m\ge0.50\), or \((0.80)^m\le0.50\). 2. This gives \(m\ge\frac{\log(0.50)}{\log(0.80)}\approx3.11\), so the smallest integer is \(m=4\). 3. The cumulative probability at \(4\) is \(1-(0.80)^4=0.5904\), while the cumulative probability at \(3\) is below \(0.50\). 4. \(P(X=4)=(0.80)^3(0.20)=0.1024\).

Answer

a) The median is \(m=4\). b) \(P(X=m)=0.1024\).

Geometric distribution 45

A geometric process averages \(3.5\) failures before the first success. Let \(X\) count trials through the first success. a) Find the success probability \(p\). b) Find \(P(X\le4)\). Give exact expressions and decimal approximations rounded to four places.

Hints

- Does the given mean count the successful attempt or only the failures? - Recover the trial success rate from the mean before finding any probabilities. - How can you use the complement to find the chance of success within four trials?

Solution

1. The expected number of failures before success is \(\frac{1-p}{p}=3.5\). 2. Solving \(1-p=3.5p\) gives \(p=\frac{2}{9}\approx0.2222\). 3. The failure probability is \(\frac{7}{9}\), so \(P(X\le4)=1-\left(\frac{7}{9}\right)^4=\frac{4160}{6561}\approx0.6340\).

Answer

a) \(p=\frac{2}{9}\approx0.2222\). b) \(P(X\le4)=\frac{4160}{6561}\approx0.6340\).

Geometric distribution 46

Two detectors scan once per round. In each round, detector A finds a signal with probability \(0.20\), and detector B finds it with probability \(0.30\). Their outcomes are independent of each other and across rounds. Let \(M\) be the first round in which at least one detector finds the signal. a) Explain why \(M\) is geometric and find its success probability. b) Find \(E(M)\), rounded to four decimal places. c) Find \(P(M=3)\), rounded to four decimal places.

Hints

- Define success for one round of both detectors. - What is the chance that neither detector finds the signal in a round? - What lets you use a geometric model across the successive rounds?

Solution

1. A round has no success only if both detectors fail, with probability \((0.80)(0.70)=0.56\). 2. Each round therefore succeeds with probability \(1-0.56=0.44\). Rounds are independent with the same success probability, so \(M\) is geometric. 3. The mean is \(E(M)=\frac{1}{0.44}\approx2.2727\) rounds. 4. The probability of first success on round \(3\) is \((0.56)^2(0.44)=0.137984\approx0.1380\).

Answer

a) \(M\) is geometric because rounds are independent and each has success probability \(0.44\). b) \(E(M)\approx2.2727\) rounds. c) \(P(M=3)\approx0.1380\).

Geometric distribution 47

A rare-event monitor succeeds independently on each cycle with probability \(0.15\). Let \(X\) be the cycle of the first success. a) Find the mean of \(X\). Give the exact value and a decimal approximation to two decimal places. b) Find the \(90\)th percentile, defined as the smallest integer \(n\) for which \(P(X\le n)\ge0.90\). c) Explain why the percentile is much larger than the mean.

Hints

- Treat the long-run average and the percentile as different summaries. - Use a cumulative probability inequality for the percentile. - Connect the comparison to the shape of a waiting-time distribution.

Solution

1. The mean is \(E(X)=\frac{1}{0.15}=\frac{20}{3}\approx6.67\) cycles. 2. Solve \(1-(0.85)^n\ge0.90,\) which gives \(n\ge\frac{\log(0.10)}{\log(0.85)}\approx14.17.\) 3. The smallest integer is \(n=15\), and \(P(X\le15)=1-(0.85)^{15}\approx0.9126.\) 4. The geometric distribution is right-skewed, so a relatively small proportion of long waits extends the upper percentiles far beyond the mean.

Answer

a) \(E(X)=\frac{20}{3}\approx6.67\) cycles. b) The \(90\)th percentile is \(15\) cycles. c) The long right tail makes high percentiles substantially larger than the mean.

Geometric distribution 48

For a geometric random variable \(X\) that counts trials through the first success, \(\frac{P(X>10)}{P(X>6)}=0.30\). Find the trial success probability \(p\).

Hints

- Express both tail probabilities using the same failure rate. - Simplify the ratio before solving. - Convert the recovered failure rate to the requested success rate.

Solution

1. For a geometric variable, \(P(X>k)=(1-p)^k\). 2. Therefore, \(\frac{P(X>10)}{P(X>6)}=(1-p)^4=0.30\). 3. Thus, \(1-p=0.30^{1/4}\), so \(p=1-0.30^{1/4}\approx0.2599\).

Answer

\(p=1-0.30^{1/4}\approx0.2599\).

Geometric distribution 49

On each independent trial, a component passes only if it passes both an electrical check and a mechanical check. The checks are independent within a trial, with pass probabilities \(0.60\) and \(0.70\). Let \(X\) be the trial number of the first component that passes both checks. a) Find the geometric success probability for \(X\). b) Find \(E(X)\). Give the exact value and a decimal approximation to three decimal places. c) Find \(P(X>3)\). Give the exact decimal value.

Hints

- Define success for one complete trial before using a waiting-time model. - Use the relationship between the two checks within a trial. - Interpret a wait beyond three trials as three combined-trial failures.

Solution

1. A trial succeeds only when both checks pass, so \(p=(0.60)(0.70)=0.42.\) 2. The mean is \(E(X)=\frac1{0.42}=\frac{50}{21}\approx2.381\) trials. 3. The failure probability is \(0.58\), so \(P(X>3)=(0.58)^3=0.195112.\)

Answer

a) \(p=0.42\). b) \(E(X)=\frac{50}{21}\approx2.381\) trials. c) \(P(X>3)=0.195112\).

Geometric distribution 50

A geometric random variable \(X\) counts trials through the first success, where \(0<p<1\). Suppose \(P(X>7)=P(X=7)\). Find the success probability \(p\).

Hints

- Write the tail and point probabilities using the same success rate. - Cancel the common nonzero factor shared by both sides. - Check that the resulting value lies in the stated interval.

Solution

1. The tail probability is \(P(X>7)=(1-p)^7\). 2. The point probability is \(P(X=7)=(1-p)^6p\). 3. Since \(0<p<1\), divide by \((1-p)^6\) to obtain \(1-p=p\). 4. Therefore, \(p=\frac{1}{2}\).

Answer

\(p=\frac{1}{2}\).

Geometric distribution 51

A geometric random variable \(X\) counts trials through the first success, with \(p=0.30\). Given that the first trial failed, find the conditional mean and standard deviation of \(X\). Give exact values and decimal approximations to three decimal places.

Hints

- Reframe the process after the known failed trial. - Separate the already-used trial from the remaining random wait. - A fixed shift affects center and spread differently.

Solution

1. Given the first failure, the remaining number of trials through success has the original geometric distribution by memorylessness. 2. Thus, conditional on \(X>1\), \(X\) has the same distribution as \(1+X'\), where \(X'\) is geometric with \(p=0.30\). 3. The conditional mean is \(E(X\mid X>1)=1+\frac{1}{0.30}=1+\frac{10}{3}=\frac{13}{3}\approx4.333\). 4. Adding \(1\) does not change spread, so \(\sigma_{X\mid X>1}=\frac{\sqrt{0.70}}{0.30}=\frac{\sqrt{70}}{3}\approx2.789\).

Answer

\(E(X\mid X>1)=\frac{13}{3}\approx4.333\) trials, and \(\sigma_{X\mid X>1}=\frac{\sqrt{70}}{3}\approx2.789\) trials.

Geometric distribution 52

Ten independent devices are each tested repeatedly until their first success. Every test succeeds with probability \(0.30\), independently across tests and devices. Let \(X_i\) be the test count for device \(i\). Find the probability that at least one device requires more than \(8\) tests.

Hints

- Start with the tail probability for one device. - Use the complement of the event involving at least one device. - Apply independence across the ten device-level events.

Solution

1. For one device, \(P(X_i>8)=(0.70)^8=0.05764801\). 2. The probability that one device succeeds within \(8\) tests is \(1-(0.70)^8\). 3. Independence across devices gives the probability all ten succeed within \(8\) tests as \([1-(0.70)^8]^{10}\). 4. Therefore, the requested probability is \(1-[1-(0.70)^8]^{10}\approx0.4478\).

Answer

\(1-[1-(0.70)^8]^{10}\approx0.4478\).

Geometric distribution 53

A technician can search for a target using either method below. Within each method, attempts are independent and the chance of success stays the same. Each attempt takes a fixed amount of time and counts toward a deadline only if it is completed. The quick scan takes \(4\,\text{s}\) per attempt and succeeds with probability \(0.20\). The detailed scan takes \(7\,\text{s}\) per attempt and succeeds with probability \(0.30\). a) Compare the expected search times, rounded to two decimal places. b) For each method, find the probability of finding the target within \(20\,\text{s}\), rounded to four decimal places. c) Which method is better by each criterion?

Hints

- How can you convert an expected number of attempts to expected seconds? - Count only attempts that finish within the deadline for each method. - Could it be easier to calculate the chance of no success before the deadline?

Solution

1. The quick scan needs an average of \(\frac{1}{0.20}=5\) attempts, so its expected time is \(4\cdot5=20\,\text{s}\). 2. The detailed scan needs an average of \(\frac{1}{0.30}=\frac{10}{3}\) attempts, so its expected time is \(\frac{70}{3}\approx23.33\,\text{s}\). 3. In \(20\,\text{s}\), the quick scan can finish \(5\) attempts, giving success probability \(1-(0.80)^5=0.67232\approx0.6723\). 4. The detailed scan can finish only \(2\) attempts within \(20\,\text{s}\), giving \(1-(0.70)^2=0.51=0.5100\). 5. The quick scan is better under both criteria.

Answer

a) Quick scan: \(20.00\,\text{s}\); detailed scan: approximately \(23.33\,\text{s}\). b) Quick scan: \(0.6723\); detailed scan: \(0.5100\). c) The quick scan is better for both expected time and success within the deadline.

Geometric distribution 54

A geometric random variable \(X\) counts trials through the first success and has variance \(6\). Find the success probability \(p\) and the mean of \(X\).

Hints

- Relate the given spread measure to the unknown trial success rate. - Expect an algebraic equation with more than one root. - Check candidate roots against the constraints on a probability.

Solution

1. For a geometric variable, \(\operatorname{Var}(X)=\frac{1-p}{p^2}\). 2. Set \(\frac{1-p}{p^2}=6\), giving \(6p^2+p-1=0\). 3. The roots are \(p=\frac{1}{3}\) and \(p=-\frac{1}{2}\); only \(p=\frac{1}{3}\) is a valid probability. 4. The mean is \(E(X)=\frac{1}{p}=3\) trials.

Answer

\(p=\frac{1}{3}\), and \(E(X)=3\) trials.

Geometric distribution 55

A laboratory allows at most \(4\) attempts to obtain a valid reading. Each attempt succeeds independently with probability \(0.25\). Let \(N\) be the number of attempts actually used; if the first three attempts fail, then \(N=4\) whether or not the fourth succeeds. a) Construct the probability distribution of \(N\). b) Find \(E(N)\).

Hints

- Separate stopping early from reaching the final allowed attempt. - Notice that the last probability includes two possible outcomes on the last attempt. - Use the completed distribution to compute the long-run average attempts used.

Solution

1. For \(N=1,2,3\), the process stops with the first success, giving probabilities \(0.25\), \((0.75)(0.25)=0.1875\), and \((0.75)^2(0.25)=0.140625\). 2. The event \(N=4\) occurs whenever the first three attempts fail, so \(P(N=4)=(0.75)^3=0.421875\). 3. The expected value is \(E(N)=1(0.25)+2(0.1875)+3(0.140625)+4(0.421875)=\frac{175}{64}\approx2.734\).

Answer

a) <table> <tr><th>\(n\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><th>\(P(N=n)\)</th><td>\(0.25\)</td><td>\(0.1875\)</td><td>\(0.140625\)</td><td>\(0.421875\)</td></tr> </table> b) \(E(N)=\frac{175}{64}\approx2.734\) attempts.

Geometric distribution 56

A geometric random variable \(X\) counts attempts through the first success, with success probability \(0.30\). Given that success occurs within the first \(5\) attempts, find the conditional expected attempt number \(E(X\mid X\le5)\). Round to three decimal places.

Hints

- Which attempt numbers remain possible once you apply the condition? - How must their probabilities change when you restrict attention to those attempts? - Form the weighted sum for only the allowed outcomes, then account for the probability of the condition.

Solution

1. The conditioning probability is \(P(X\le5)=1-(0.70)^5=0.83193\). 2. The weighted probability sum over the allowed values is \(\sum_{k=1}^{5}k(0.70)^{k-1}(0.30)=1.93275\). 3. Therefore, \(E(X\mid X\le5)=\frac{1.93275}{0.83193}\approx2.323\) attempts.

Answer

\(E(X\mid X\le5)\approx2.323\) attempts.

Geometric distribution 57

A researcher can run up to \(3\) independent searches each day. Each search finds the target with probability \(0.20\), independently of searches on other days. The researcher stops for the day after success; if all three searches fail, the process resumes the next day under the same conditions. Let \(D\) be the day of the first success. a) Explain why \(D\) has a geometric distribution and find its daily success probability. b) Find \(P(D=2)\), rounded to four decimal places. c) Find \(E(D)\), rounded to three decimal places.

Hints

- Treat the whole day's searches as one repeated trial. - What is the chance that all searches in a day fail? - Once you have the day-level success probability, which waiting-time formulas apply?

Solution

1. A day is successful when at least one of its three searches succeeds. Days repeat with independent, identical search conditions. 2. The daily success probability is \(r=1-(0.80)^3=0.488\), so \(D\) has a geometric distribution with parameter \(0.488\). 3. \(P(D=2)=(1-r)r=(0.512)(0.488)=0.249856\approx0.2499\). 4. \(E(D)=\frac{1}{0.488}\approx2.049\) days.

Answer

a) \(D\) is geometric because days form independent trials with the same chance of at least one success. That chance is \(0.488\). b) \(P(D=2)\approx0.2499\). c) \(E(D)\approx2.049\) days.

Geometric distribution 58

A geometric random variable \(X\) counts trials through the first success, with \(p=0.40\). a) Find the probability that \(X\) is odd. b) Find the probability that \(X\) is even.

Hints

- Write the first few point probabilities for one parity. - Identify the common multiplier between consecutive terms in that list. - Use the fact that every positive integer is either odd or even.

Solution

1. Odd values are \(1,3,5,\ldots\), so their probabilities form \(0.40+(0.60)^2(0.40)+(0.60)^4(0.40)+\cdots\). 2. This geometric series has sum \(\frac{0.40}{1-(0.60)^2}=\frac{5}{8}=0.625\). 3. The even probability is the complement, \(1-\frac{5}{8}=\frac{3}{8}=0.375\).

Answer

a) \(P(X\text{ is odd})=\frac{5}{8}=0.625\). b) \(P(X\text{ is even})=\frac{3}{8}=0.375\).

Geometric distribution 59

Independent geometric random variables \(X\) and \(Y\) count trials through first success, with success probabilities \(0.20\) and \(0.30\), respectively. Let \(T=\max(X,Y)\), the number of rounds needed until both processes have succeeded at least once. Find \(P(T\le4)\). Is \(T\) itself geometric? Explain.

Hints

- Rewrite the maximum event as simultaneous conditions on the two component waits. - Use the independence of the two processes after finding each cumulative probability. - For the model question, examine whether the state of the process is unchanged after every unfinished round.

Solution

1. The event \(T\le4\) occurs when both \(X\le4\) and \(Y\le4\). 2. Independence gives \(P(T\le4)=[1-(0.80)^4][1-(0.70)^4]\approx0.4486\). 3. The chance that the process finishes in a round depends on whether neither, one, or both component processes have already succeeded. Therefore, \(T\) does not have a constant success probability from round to round and is not geometric.

Answer

\(P(T\le4)=[1-(0.80)^4][1-(0.70)^4]\approx0.4486\). The variable \(T\) is not geometric.

Geometric distribution 60

A geometric random variable \(X\) counts trials through the first success. It is known that \(P(X=2\mid X\le2)=0.40\). Find the success probability \(p\).

Hints

- Write the exact and cumulative probabilities using the same failure rate. - Simplify the conditional ratio before solving. - Convert the resulting failure probability back to a success probability.

Solution

1. Let \(q=1-p\). Then \(P(X=2)=qp\) and \(P(X\le2)=p+qp=p(1+q)\). 2. The conditional probability is \(\frac{q}{1+q}=0.40\). 3. Solving gives \(q=\frac{2}{3}\), so \(p=1-q=\frac{1}{3}\).

Answer

\(p=\frac{1}{3}\).

Geometric distribution 61

Before a sequence of trials begins, a hidden setting is chosen with equal probability. Under the low setting, every trial succeeds with probability \(0.20\); under the high setting, every trial succeeds with probability \(0.80\). Conditional on the chosen setting, trials are independent. Let \(X\) be the trial of the first success. a) Find \(P(X=3)\). b) Explain why using a geometric model with the average success probability \(0.50\) is incorrect.

Hints

- Condition on each possible setting before combining probabilities. - Keep the setting fixed throughout a sequence of trials. - Compare the resulting mixture with what a single constant-rate model would predict.

Solution

1. Under the low setting, \(P(X=3)=(0.80)^2(0.20)=0.128\). 2. Under the high setting, \(P(X=3)=(0.20)^2(0.80)=0.032\). 3. Averaging over the equally likely settings gives \(P(X=3)=0.5(0.128)+0.5(0.032)=0.080\). 4. A geometric model with \(p=0.50\) would give \((0.50)^2(0.50)=0.125\). The shared hidden setting makes trial outcomes dependent when the setting is not observed, so averaging the two success rates does not create a geometric process.

Answer

a) \(P(X=3)=0.080\). b) The hidden setting persists across trials, creating dependence in the unconditional process. A geometric model with \(p=0.50\) would incorrectly give \(0.125\).

Geometric distribution 62

A geometric random variable \(X\) counts trials through the first success, with \(p=0.25\). Find the probability that the first success occurs on a trial number divisible by \(4\).

Hints

- List the trial numbers that meet the divisibility condition. - Write the first few point probabilities and identify their common ratio. - Sum the repeating infinite pattern.

Solution

1. The possible values are \(4,8,12,\ldots\). 2. Their probabilities form \((0.75)^3(0.25)+(0.75)^7(0.25)+(0.75)^{11}(0.25)+\cdots\). 3. This is a geometric series with first term \((0.75)^3(0.25)\) and common ratio \((0.75)^4\). 4. The sum is \(\frac{(0.75)^3(0.25)}{1-(0.75)^4}=\frac{27}{175}\approx0.1543\).

Answer

\(P(X\text{ is divisible by }4)=\frac{27}{175}\approx0.1543\).

Geometric distribution 63

A geometric random variable \(X\) counts trials through the first success, with \(p=0.30\). Given that \(X\) is odd, find \(P(X=3\mid X\text{ is odd})\).

Hints

- What is the probability of first success on the third attempt? - The condition includes infinitely many possible waiting times. What pattern links their probabilities? - How should the probability of \(X=3\) be adjusted after restricting attention to odd waits?

Solution

1. The point probability is \(P(X=3)=(0.70)^2(0.30)=0.147\). 2. The probability of an odd waiting time is \(0.30+(0.70)^2(0.30)+(0.70)^4(0.30)+\dots=\frac{0.30}{1-(0.70)^2}=\frac{10}{17}\). 3. Therefore, \(P(X=3\mid X\text{ is odd})=\frac{0.147}{\frac{10}{17}}=0.2499\).

Answer

\(P(X=3\mid X\text{ is odd})=0.2499\).

Geometric distribution 64

A geometric random variable \(X\) counts trials through the first success, where \(0<p<1\). Suppose \(P(X\le3)=2P(X=1)\). Find \(p\).

Hints

- Express the cumulative and first-trial probabilities using the failure rate. - Look for a factorization before solving the resulting equation. - Retain only a root that represents a probability.

Solution

1. Let \(q=1-p\). The condition becomes \(1-q^3=2p=2(1-q)\). 2. Factor \(1-q^3=(1-q)(1+q+q^2)\). Since \(q\ne1\), cancel \(1-q\) to get \(1+q+q^2=2\). 3. Thus, \(q^2+q-1=0\), and the valid root is \(q=\frac{\sqrt{5}-1}{2}\). 4. Therefore, \(p=1-q=\frac{3-\sqrt{5}}{2}\approx0.3820\).

Answer

\(p=\frac{3-\sqrt{5}}{2}\approx0.3820\).

Geometric distribution 65

For a geometric random variable \(X\) that counts trials through the first success, the ratio of the standard deviation to the mean is \(0.80\). Find the success probability \(p\) and the mean of \(X\). Give the mean exactly and to four decimal places.

Hints

- Form the requested ratio before substituting the given value. - Simplify common factors in the two distribution summaries. - Convert the resulting failure probability to a success probability.

Solution

1. For a geometric variable, \(\frac{\sigma_X}{E(X)} =\frac{\sqrt{1-p}/p}{1/p} =\sqrt{1-p}.\) 2. Thus, \(\sqrt{1-p}=0.80,\) so \(1-p=0.64\) and \(p=0.36\). 3. The mean is \(E(X)=\frac{1}{0.36}=\frac{25}{9}\approx2.7778\) trials.

Answer

\(p=0.36\), and \(E(X)=\frac{25}{9}\approx2.7778\) trials.

Geometric distribution 66

A system succeeds independently on each attempt with probability \(0.25\). Before testing, an attempt limit \(L\) is chosen uniformly from \(1,2,3,4,5\), independently of all test outcomes. Testing stops at the first success or after the last allowed attempt. Find the probability of at least one success by the final allowed attempt, including success on that last attempt. Give an exact value and a decimal rounded to four places.

Hints

- First imagine the maximum number of attempts is fixed. What is the chance of no success by that limit? - How does the chance of success change with the chosen limit? - How should the probabilities for the five equally likely limits be combined?

Solution

1. Conditional on \(L=\ell\), the probability of at least one success is \(1-(0.75)^\ell\). 2. The five limits are equally likely, so the probability is \(\frac{1}{5}\sum_{\ell=1}^{5}\left[1-\left(\frac{3}{4}\right)^\ell\right]\). 3. This equals \(\frac{2777}{5120}=0.5423828125\approx0.5424\).

Answer

\(\frac{2777}{5120}\approx0.5424\).

Geometric distribution 67

A decoding process succeeds independently with probability \(0.50\) on each attempt. Attempt \(1\) takes \(1\) second, attempt \(2\) takes \(2\) seconds, and in general attempt \(k\) takes \(k\) seconds. Let \(X\) be the attempt of the first success and \(T\) the total elapsed time through that attempt. Find \(E(T)\).

Hints

- Express the accumulated time in terms of the random attempt number. - Rewrite the expected value using moments of the waiting-time variable. - Use the relationship between variance, the mean, and the second moment.

Solution

1. If success occurs on attempt \(X\), then \(T=1+2+\cdots+X=\frac{X(X+1)}{2}\). 2. For \(p=0.50\), \(E(X)=2\) and \(\operatorname{Var}(X)=\frac{0.50}{0.50^2}=2\). 3. Therefore, \(E(X^2)=\operatorname{Var}(X)+[E(X)]^2=2+4=6\). 4. Thus, \(E(T)=\frac{E(X^2)+E(X)}{2}=\frac{6+2}{2}=4\) seconds.

Answer

\(E(T)=4\) seconds.

Geometric distribution 68

Two processes begin together and each is attempted once per round. Process A succeeds with probability \(0.40\) per round, and process B succeeds with probability \(0.25\) per round. Results are independent between the processes and from round to round. Let \(X\) and \(Y\) be the rounds of their first successes. Find \(P(X<Y)\), \(P(X=Y)\), and \(P(Y<X)\).

Hints

- List all the ways a round can end: neither succeeds, A alone succeeds, B alone succeeds, or both succeed. - Before a decisive round, which earlier outcome can repeat any number of times? - What should the probabilities of the three final comparisons add up to?

Solution

1. A round in which neither process succeeds has probability \((0.60)(0.75)=0.45\). 2. The first decisive round favors A if A succeeds while B fails, with probability \((0.40)(0.75)=0.30\). Any number of earlier double failures is possible, so \(P(X<Y)=\frac{0.30}{1-0.45}=\frac{6}{11}\). 3. A tie in the first decisive round has probability \((0.40)(0.25)=0.10\), giving \(P(X=Y)=\frac{0.10}{0.55}=\frac{2}{11}\). 4. A win for B in the first decisive round has probability \((0.60)(0.25)=0.15\), giving \(P(Y<X)=\frac{0.15}{0.55}=\frac{3}{11}\). 5. The three probabilities add to \(1\), as they must.

Answer

\(P(X<Y)=\frac{6}{11}\), \(P(X=Y)=\frac{2}{11}\), and \(P(Y<X)=\frac{3}{11}\).

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