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### Geometric Probability Distribution

A probability distribution for the random variable [tex]$X$[/tex], 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 356), except that the number of trials is not fixed. Its probability distribution function (pdf) is:

[tex]\[
P(x) = p(1-p)^{x-1}, \quad x = 1, 2, 3, \ldots
\][/tex]

where [tex]\( p \)[/tex] 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 [tex]\( 52.4\% \)[/tex] of his free throws. That is, find [tex]\( P(3) \)[/tex].

(b) Construct a probability distribution for the random variable [tex]\( X \)[/tex], the number of free-throw attempts of Shaquille O'Neal until he makes a free throw. Construct the distribution for [tex]\( x = 1, 2, 3, \ldots, 10 \)[/tex]. The probabilities are small for [tex]\( x \ \textgreater \ 10 \)[/tex].

(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 [tex]\( \frac{1}{p} \)[/tex]. Conclude that the mean of a geometric probability distribution is the expected number of attempts Shaquille O'Neal will take before we observe a made free throw.



Answer :

GinnyAnswer
Alright, let's go through the geometric probability distribution problem step-by-step.

### (a) Probability [tex]\( P(3) \)[/tex]

Given:
- [tex]\( p = 0.524 \)[/tex] (Shaquille O'Neal's success rate)
- [tex]\( x = 3 \)[/tex] (we want the probability he makes his third attempt after missing two)

The probability [tex]\( P(x) \)[/tex] for a geometric distribution is given by:
[tex]\[ P(x) = p (1 - p)^{x-1} \][/tex]

So, for [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^{3-1} \][/tex]

From the result obtained:
[tex]\[ P(3) = 0.118725824 \][/tex]

### (b) Probability Distribution for [tex]\( x = 1, 2, \ldots, 10 \)[/tex]

We need to construct a probability distribution for [tex]\( x = 1 \)[/tex] to [tex]\( x = 10 \)[/tex]. The probability distribution [tex]\( P(x) \)[/tex] is:

1. For [tex]\( x = 1 \)[/tex]:
[tex]\[ P(1) = 0.524 \][/tex]

2. For [tex]\( x = 2 \)[/tex]:
[tex]\[ P(2) = 0.524 \times (1 - 0.524)^1 = 0.249424 \][/tex]

3. For [tex]\( x = 3 \)[/tex]:
[tex]\[ P(3) = 0.524 \times (1 - 0.524)^2 = 0.118725824 \][/tex]

4. For [tex]\( x = 4 \)[/tex]:
[tex]\[ P(4) = 0.524 \times (1 - 0.524)^3 = 0.056513492224 \][/tex]

5. For [tex]\( x = 5 \)[/tex]:
[tex]\[ P(5) = 0.524 \times (1 - 0.524)^4 = 0.026900422298624 \][/tex]

6. For [tex]\( x = 6 \)[/tex]:
[tex]\[ P(6) = 0.524 \times (1 - 0.524)^5 = 0.012804601014145 \][/tex]

7. For [tex]\( x = 7 \)[/tex]:
[tex]\[ P(7) = 0.524 \times (1 - 0.524)^6 = 0.006094990082733 \][/tex]

8. For [tex]\( x = 8 \)[/tex]:
[tex]\[ P(8) = 0.524 \times (1 - 0.524)^7 = 0.002901215279381 \][/tex]

9. For [tex]\( x = 9 \)[/tex]:
[tex]\[ P(9) = 0.524 \times (1 - 0.524)^8 = 0.001380978472985 \][/tex]

10. For [tex]\( x = 10 \)[/tex]:
[tex]\[ P(10) = 0.524 \times (1 - 0.524)^9 = 0.000657345753141 \][/tex]

Thus, the probability distribution is:
[tex]\[ [0.524, 0.249424, 0.118725824, 0.056513492224, 0.026900422298624, 0.012804601014145, 0.006094990082733, 0.002901215279381, 0.001380978472985, 0.000657345753141] \][/tex]

### (c) Mean of the Distribution

The mean [tex]\( \mu \)[/tex] of a geometric distribution is given by:
[tex]\[ \mu = \frac{1}{p} \][/tex]

Using [tex]\( p = 0.524 \)[/tex]:
[tex]\[ \mu = \frac{1}{0.524} \][/tex]

From the result obtained:
[tex]\[ \mu = 1.9083969465648853 \][/tex]

### (d) Comparison of the Mean with [tex]\( \frac{1}{p} \)[/tex]

We have already computed the mean as [tex]\( \frac{1}{p} = 1.9083969465648853 \)[/tex].

So, the mean calculated in part (c) [tex]\( \mu = 1.9083969465648853 \)[/tex] is indeed equal to [tex]\( \frac{1}{p} \)[/tex]. This confirms that the mean of a geometric distribution is [tex]\( \frac{1}{p} \)[/tex].

In conclusion, the expected number of free throw attempts Shaquille O'Neal would make before observing a successful free throw is approximately 1.908.