Section 2.3

Discrete Random Variables and Distributions

Part 1: Discrete Distributions

Question 2.3.1

Suppose that a symmetrical die is rolled 20 independent times, and each time we record whether or not the event \(\{2, 3, 5, 6\}\) has occurred.

  1. What is the distribution of the number of times this event occurs in 20 rolls?

  2. Calculate the probability that the event occurs five times.

\(\text{Binomial}(n = 20, \theta =2/3)\)

\[ \mathbb{P}(X = 5) = \binom{20}{5}(2/3)^5(1/3)^{15} = 1.4229 \times 10^{-4} \]

Question 2.3.2

An urn contains 4 black balls and 5 white balls. After a thorough mixing, a ball is drawn from the urn, its color is noted, and the ball is set aside. The remaining balls are then mixed and a second ball is drawn.

  1. What is the probability distribution of the number of black balls observed?

  2. What is the probability distribution of the number of white balls observed?

\(\text{Hypergeometric}(N = 9, M= 4, k = 2)\)

\(\text{Hypergeometric}(N = 9, M = 5, k = 2)\)

Part 2 (Discrete Families)

Question 2.3.3

Let \(Z \sim \text{Geometric}(\theta = 0.4)\). Compute \(\mathbb{P}(7 \leq Z < 9)\).

\[ \begin{aligned} \mathbb{P}(7 \leq Z < 9) &= \mathbb{P}(Z = 7) + \mathbb{P}(Z = 8) \\ &= (0.4)(1-0.4)^7+(0.4)(1-0.4)^8\\ &= 0.0179 \end{aligned} \]

Question 2.3.4

Let \(X \sim \text{Binomial}(12, \theta)\). For what value of \(\theta\) is \(\mathbb{P}(X = 11)\) maximized?

\(P(X = 11) = {12\choose 11}\theta^{11}(1-\theta)^1 = 12\theta^{11}(1-\theta)\)

To find the inflection point(s), we can take the derivative and set it equal to zero.

\[ \begin{aligned} \frac{d}{d\theta}12\theta^{11}(1-\theta) &=\left[\frac{d}{d\theta}12\theta^{11}\right](1-\theta) + 12\theta^{11}\left[\frac{d}{d\theta}(1-\theta)\right] \\ &= (12)(11)\theta^{10}(1-\theta) + 12\theta^{11}(-1) \end{aligned} \] Then, \[ \begin{aligned} &\hspace{1.5cm} (12)(11)\theta^{10}(1-\theta) + 12\theta^{11}(-1) = 0\\ &\implies 11\theta^{10}(1-\theta) - 12\theta^{11} = 0\\ &\implies 11\theta^{10}(1-\theta) = 12\theta^{11} \end{aligned} \] This will be zero when \(\theta = 0\). We can also consider the case of \(\theta \ne 0\): \[ \begin{aligned} &\implies 11(1-\theta) = \theta && \text{if } \theta \ne 0 \text{ can divide by } \theta^{10}\\ &\implies 11-11\theta = \theta \\ &\implies 11 = 12\theta \\ &\implies \theta = 11/12 \\ \end{aligned} \] We need to check that this is a maximum using second derivative test:

\[ \begin{aligned} \frac{d^2}{d\theta^2}12\theta^{11}(1-\theta) &= \frac{d}{d\theta}\left[(12)(11)\theta^{10}(1-\theta) - 12\theta^{11}\right]\\ &= \frac{d}{d\theta}\left[132\theta^{10}-132\theta^{11} - 12\theta^{11}\right]\\ &= 132(10)\theta^{9}-132(11)\theta^{10} - 12(11)\theta^{10} \end{aligned} \] We can plug the critical points into here and see if it is negative. If it is, then it’s a maximum:

The second derivative evaluated at \(\theta=0\) is 0, indicating it is a saddle point or local max/min, (not a maximum).

The second derivative evaluated at \(\theta=11/12\) is: \[ \begin{aligned} 132(10)(11/12)^{9}-132(11)(11/12)^{10} - 12(11)(11/12)^{10} \approx -60.32. \end{aligned} \] which indicates that this point is a maximum.

Question 2.3.5

A mystery novel book club has 20 members. 8 of them have secretly already read this month’s selection, and know the ending. The organizer randomly calls on 6 members during the meeting to share their predictions how the novel ends. What is the probability that no more than 2 of the 6 people know how the story ends?

This distribution is the Hypergeometric distribution with \(N\) = 20, \(K = 8\), \(n\) = 6.

\[ \begin{aligned} \mathbb{P}(X \le 2) &= \mathbb{P}(X = 0) + \mathbb{P}(X = 1) + \mathbb{P}(X = 2)\\ &= \frac{{8 \choose 0} {12\choose 6}}{{20\choose 6}} + \frac{{8 \choose 1}{12\choose 5}}{{20\choose 6}} + \frac{{8 \choose 2} {12\choose 4}}{{20\choose 6}} \\ &= 0.0238 + 0.1634 + 0.3581\\ &= 0.5453 \end{aligned} \]

Question 2.3.6

A small coastal town’s lighthouse keeper records the number of ships passing by the lighthouse. On average, 3 ships pass by per hour, and the number of ships in any given hour follows a Poisson distribution.

  1. What is the probability that exactly 5 ships pass by during a given hour?

  2. What is the probability that no ships pass by during a given hour?

  3. What is the probability that at least 2 ships pass by during a given hour?

Let \(X\) be the number of ships passing by in one hour. Then \(X \sim \text{Poisson}(\lambda = 3)\), since the average rate is 3 ships per hour.

We want the probability of observing exactly 5 ships: \[ \begin{aligned} \mathbb{P}(X = 5) &= \frac{e^{-3}(3)^5}{5!} \\ &= \frac{0.049787 \times 243}{120} \\ &= \frac{12.0982}{120} \\ &\approx 0.1008 \end{aligned} \] Therefore, the probability that exactly 5 ships pass by during a given hour is \(\boldsymbol{\approx 0.1008}\).

We want the probability of observing zero ships: \[ \begin{aligned} \mathbb{P}(X = 0) &= \frac{e^{-3}(3)^0}{0!} \\ &= e^{-3} \\ &\approx 0.0498 \end{aligned} \] Therefore, the probability that no ships pass by during a given hour is \(\boldsymbol{\approx 0.0498}\).

“At least 2 ships” is the complement of observing 0 or 1 ships, so \[ \mathbb{P}(X \geq 2) = 1 - \mathbb{P}(X = 0) - \mathbb{P}(X = 1) \] We already found \(\mathbb{P}(X = 0) \approx 0.0498\) in part (b). We also need \[ \begin{aligned} \mathbb{P}(X = 1) &= \frac{e^{-3}(3)^1}{1!} \\ &= 3e^{-3} \\ &\approx 0.1494 \end{aligned} \] So, \[ \begin{aligned} \mathbb{P}(X \geq 2) &= 1 - 0.0498 - 0.1494 \\ &\approx 0.8008 \end{aligned} \] Therefore, the probability that at least 2 ships pass by during a given hour is \(\boldsymbol{\approx 0.8008}\).

Question 2.3.7

A salesperson makes independent phone calls to potential customers. On each call, the probability of making a sale is 0.25.

  1. What is the probability that the salesperson makes their first sale on the 4th call?

  2. What is the probability that it takes more than 3 calls to make the first sale?

Let \(X\) be the number fails before the first successful sale. Then \(X \sim \text{Geometric}(\theta = 0.25)\), since each call is an independent trial with probability of success \(\theta = 0.25\).

We want the probability that the first sale occurs on exactly the 4th call, which requires 3 failures followed by a success: \[ \begin{aligned} \mathbb{P}(X = 3) &= (1-\theta)^{3}\,\theta \\ &= (0.75)^3(0.25) \\ &= (0.421875)(0.25) \\ &\approx 0.1055 \end{aligned} \] Therefore, the probability that the salesperson makes their first sale on the 4th call is \(\boldsymbol{\approx 0.1055}\).

“More than 3 calls” is the event \(X > 3\), i.e., the first 3 calls are all failures: \[ \begin{aligned} \mathbb{P}(X > 3) &= (1-\theta)^3 \\ &= (0.75)^3 \\ &= 0.421875 \end{aligned} \] Therefore, the probability that it takes more than 3 calls to make the first sale is \(\boldsymbol{0.421875}\).

Question 2.3.8

A basketball player makes free throws independently, with probability 0.6 of success on each attempt. What is the probability that the player makes their 3rd successful free throw on their 7th attempt?

Let \(X\) be the number of failures before the 3rd successful free throw. Then \(X \sim \text{NegativeBinomial}(r = 3, \theta = 0.6)\), since we are counting the number of independent trials needed to observe the 3rd success, with probability of success \(\theta = 0.6\) on each trial.

To have the 3rd successful free throw on their 7th attempt, it means we need \(X = 4\) failures before the 3rd success.

\[ \begin{aligned} \mathbb{P}(X = x) &= \binom{r - 1 +x}{x}\theta^{r}(1-\theta)^{x} \\ &= \binom{3 - 1 + 4}{4}0.6^{3}(1-0.6)^{4} \\ &= \binom{6}{4}\theta^{3}(1-\theta)^{4} \\ &= 15 (0.6)^3(0.4)^4 \\ &= 15 \times 0.216 \times 0.0256 \\ &\approx 0.0829 \end{aligned} \]