Question

The Geometric Distribution. In this exercise, we discuss the geometric distribution, the probability distribution for the number of trials until the first success in Bernoulli trials. The geometric probability formula is

$P(X=x)=p{(1-p)}^{x-1},$

where Xdenotes the number of trials until the first success and pthe success probability. Using the geometric probability formula and Definition 5.9 on page 227. we can show that the mean of the random variable Xis 1/p.

To illustrate, consider the Mega Millions lottery, a multi-state jackpot draw game with a jackpot starting at $15 million and growing until someone wins. In order to play, the player selects five white numbers from the numbers 1-75 and one Mega Ball number from the numbers 1-15. Suppose that you buy one Mega Millions ticket per week. Let Xdenote the number of weeks until you win a prize.

(a) Find and interpret the probability formula for the random variable X. (Note: The probability of winning a prize with a single ticket is 0.0680.)

(b) Compute the probability that the number of weeks until you win a prize is exactly 3; at most 3: at least 3.

(c) On average, how long will it be until you win a prize?

Short answer

Part (a) 0.0290647

Part (b)

$$\begin{gathered} P\left(X=3\right)=0.027 \\ P\left(X\ge 3\right)=0.085 \\ P\left(X\le 3\right)=0.9427154 \end{gathered}$$

Part (c) approx 34 weeks.

Step-by-step solution

Part (a) Step 1. Given information.

The given statement is:

The formula of geometric probability is:

$P\left(X=x\right)=p{\left(1-p\right)}^{x-1}$

Part (a) Step 2. Find the probability formula for the random variable X.

Let's call 'X' the number of winning numbers on a single ticket:

No. of Winning Numbers	Probability
0	0.3713060
1	0.4311941
2	0.1684352
3	0.0272219
4	0.0018014
5	0.0000412
6	0.0000002

To win a prize, the ticket must include three or more of the winning numbers. That is, for winning a prize in a lottery ticket, $x\ge 3$.

Part (a) Step 3. Find the probability.

Consider X to be the number of weeks until you win a prize. That is, X can have the values 1, 2, 3, 4, and so on.

We can clearly see that, Xhas a geometric distribution with parameters p, where prepresents the probability of success.

P can be described as:

$$\begin{gathered} p=P\left(X\ge 3\right) \\ =P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right) \\ =0.0272219+0.0018014+0.0000412+0.0000002 \\ =0.0290647 \end{gathered}$$

As a result, we can say that there is only a 2.9 percent chance of winning a prize on a lottery ticket.

Part (b) Step 1. Find the probability that the number of weeks until you win a prize is exactly 3; at most 3: at least 3.

Use the formula of geometric probability:

$$\begin{gathered} P\left(X=x\right)=p{\left(1-p\right)}^{x-1} \\ P\left(X=3\right)=0.029{\left(1-0.029\right)}^{3-1} \\ =0.029\times {\left(0.971\right)}^2 \\ =0.027 \end{gathered}$$

Probability of at most 3 wins:

role="math" localid="1651849915949" $$\begin{gathered} P\left(X\le 3\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right) \\ =0.029{\left(1-0.029\right)}^{1-1}+0.029{\left(1-0.029\right)}^{2-1}+0.029{\left(1-0.029\right)}^{3-1} \\ =0.029\left(1+0.970+{\left(0.971\right)}^2\right) \\ =0.029+0.028+0.027 \\ =0.085 \end{gathered}$$

Probability of at least 3 wins:

$$\begin{gathered} P\left(X>3\right)=1-\left[P\left(X=1\right)+P\left(X=2\right)\right] \\ =1-0.029\left(1+0.970\right) \\ =1-\left(0.029+0.028\right) \\ =0.9427154 \end{gathered}$$

Part (c) Step 1. Find the time it took to win the prize.

The mean is:

$$\begin{gathered} E\left(X\right)=\frac{1}{p} \\ =\frac{1}{0.029} \\ =34.406 \\ \approx 34\mathrm{weeks} \end{gathered}$$