91Ó°ÊÓ

A probability distribution for the random variable \(X,\) the number of trials until a success is observed, is called the geometric probability distribution. It has the same criteria as the binomial distribution (see page 328 ), except that the number of trials is not fixed. Its probability distribution function (pdf) is $$ P(x)=p(1-p)^{x-1}, \quad x=1,2,3, \ldots $$ where \(p\) is the probability of success. (a) What is the probability that Shaquille O'Neal misses his first two free throws and makes the third? Over his career, he made \(52.4 \%\) of his free throws. That is, find \(P(3)\) (b) Construct a probability distribution for the random variable \(X,\) the number of free-throw attempts of Shaquille O'Neal until he makes a free throw. Construct the distribution for \(x=1,2,3, \ldots, 10 .\) The probabilities are small for \(x>10\) (c) Compute the mean of the distribution, using the formula presented in Section 6.1 (d) Compare the mean obtained in part (c) with the value \(\frac{1}{p}\). Conclude that the mean of a geometric probability distribution is \(\mu_{X}=\frac{1}{p} .\) How many free throws do we expect Shaq to take before we observe a made free throw?

Step by step solution

01

Identify Given Information

Identify the information given in the problem. Shaquille O'Neal's free throw success rate is 52.4%, which means the probability of success, denoted as \(p\), is 0.524.

02

Write the Probability Distribution Function

The probability distribution function for a geometric distribution is given by \(P(x) = p(1-p)^{x-1}\), where \(x\) is the number of trials until the first success.

03

Calculate \(P(3)\)

Substitute \(p = 0.524\) and \(x = 3\) into the probability distribution function to find the probability that Shaquille O'Neal misses his first two free throws and makes the third: \[P(3) = 0.524(1-0.524)^{3-1} = 0.524(0.476)^2\]. Compute the value.

04

Evaluate \(P(3)\)

Perform the arithmetic to evaluate \(P(3)\): \[P(3) = 0.524 \times 0.226176 = 0.11849\]. Thus, the probability that Shaquille O'Neal makes his third free throw after missing the first two is approximately 0.1185.

05

Construct the Probability Distribution for \(x = 1, 2, \ldots, 10\)

Calculate the probabilities for \(x = 1, 2, \ldots, 10\) by substituting these values of \(x\) into the probability distribution function \(P(x) = 0.524(0.476)^{x-1}\). Construct a table with these probabilities.

06

Calculate Probabilities for Each \(x\)

Using the values of \(p = 0.524\) and \(1-p = 0.476\), compute: \[P(1) = 0.524\] \[P(2) = 0.524 \times 0.476 = 0.249824\] \[P(3) = 0.524 \times 0.476^2 = 0.11849\] ... Continue computing for \(x = 4, 5, ..., 10\).

07

Compute the Mean of the Distribution

The mean of a geometric distribution is given by \(\mu_X = \frac{1}{p}\). Using \(p = 0.524\), compute: \[\mu_X = \frac{1}{0.524} = 1.908\].

08

Comparison with \(\frac{1}{p}\)

Compare the calculated mean with the theoretical mean. The calculated mean is \(1.908\), which matches \(\frac{1}{p}\). Conclude that the mean number of free throws Shaquille O'Neal is expected to take before making a successful free throw is \(1.908\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

probability distribution function

A probability distribution function (pdf) describes the likelihood of different outcomes in a random experiment. In the context of a geometric probability distribution, the pdf specifically addresses the probability of achieving the first success on the x-th trial. The formula for the geometric distribution pdf is given by: \[ P(x) = p(1-p)^{x-1} \] where:

• p is the probability of success
• (1-p) is the probability of failure
• x is the trial number on which the first success occurs

An example is given in the original exercise, where Shaquille O'Neal's free throw success rate is 52.4%. To find the probability that he makes his third shot after missing the first two, we substitute p with 0.524 and x with 3: \[ P(3) = 0.524(1-0.524)^{3-1} = 0.524(0.476)^2 \] Calculating this gives approximately 0.1185.

mean of geometric distribution

The mean of a geometric distribution is an important measure that tells us the average number of trials needed until the first success. For a geometric distribution, the mean, denoted as \( \mu_X \), is given by the formula: \[ \mu_X = \frac{1}{p} \]This formula is derived from the fact that, in a geometric setting, each trial can be seen as a Bernoulli trial, a trial with only two outcomes: success or failure. By observing the recurring nature and independence of each trial, we can determine the expected number of trials (mean) for a success to occur. In the exercise's solution, Shaquille O'Neal's free throw success rate of 52.4% translates to a mean of: \[ \mu_X = \frac{1}{0.524} \approx 1.908 \] This value means we expect Shaquille O'Neal to take, on average, about 1.908 free throws before making a successful shot. This matches the calculated mean found in part (c) of the exercise by constructing the actual probability distribution.

binomial distribution

The binomial distribution is a discrete probability distribution of the number of successes in a sequence of independent experiments with a fixed number of trials, each of which yields success with a given probability. The key characteristics of a binomial distribution are:

• Fixed number of trials
• Each trial has two possible outcomes: success or failure
• Constant probability of success in each trial
• Trials are independent of each other

The probability of getting exactly k successes in n trials is given by: \[ P(X=k) = {n \choose k} p^k (1-p)^{n-k} \]Where:

• n = total number of trials
• k = number of successes
• p = probability of success
• \( n \choose k \) = binomial coefficient

In contrast, the geometric distribution discussed in the exercise has similar criteria but without a fixed number of trials. Instead, it focuses on the number of trials needed to get the first success. This makes geometric distribution a special case of binomial distribution with the 'first success' stopping rule, without knowing the maximum number of trials in advance.