Chapter 17, Problem 5E

(a)

To determine

To Explain: the how can the number of Tiger photographs be simulated in five cereal boxes.

Answer to Problem 5E

0.0819

Explanation of Solution

Given:

  $p=0.2$

Calculation:

The probability of a faulty shot by the basketball player is 0.80. The likelihood of strong shot is then 1 - 0.80 = 0.20.

a)

It is computed as follows that the player is expected to fail his fifth attempt for the first time. $\begin{array}{c}P\left({\text{Misses for 1}}^{st}{\text{ time on 5}}^{th}\text{ attempt}\right)={\left(1-p\right)}^4\left(P\right)\\ ={\left(1-0.2\right)}^4\left(0.2\right)\\ ={0.8}^4\times 0.2\\ =0.0819\end{array}$

The probability that the player fails his fifth try for the first time is therefore 0.0819.

(b)

To determine

To Explain: the run at minimum 30 trials on the basis of given information.

Answer to Problem 5E

0.0064

Explanation of Solution

Given:

  $p=0.2$

Calculation:

On the fourth shot, the probabilities are as follows: The player will make his first basket: $\begin{array}{c}P\left({\text{ 1}}^{st}{\text{ basket on 4}}^{th}\text{ shot}\right)={\left(1-p\right)}^3\left(P\right)\\ ={\left(1-0.80\right)}^3\left(0.8\right)\\ ={0.2}^3\times 0.8\\ =0.0064\end{array}$

The probability of the player creating his first basket on his fourth attempt is therefore 0.0064.

(c)

To determine

To Calculate: the probability that get no pictures of Tiger, one photograph, two photographs etc.

Answer to Problem 5E

0.992

Explanation of Solution

Given:

  $p=0.2$

Calculation:

The chances of success the first basket is 0.8

The chances of success the second basket is $0.2\times 0.8=0.16$

The chances of success the third basket is 0 ${0.2}^2\times 0.8=0.032$

It's estimated as follows that the player is expected to make their first basket in one of their first 3 plays. $\begin{array}{c}P\left({\text{ 1}}^{st}\text{ basket in one of his 3 shots}\right)=P\left(1^{st}{\text{ basket in 1}}^{st}\text{ shot}\right)\\ +P\left(1^{st}{\text{ basket in 2}}^{nd}\text{ shot}\right)\\ +P\left(1^{st}{\text{ basket in 3}}^{ed}\text{ shot}\right)\\ =0.8+0.16+0.032\\ =0.992\end{array}$

Therefore, in one of his first 3 plays, the player would have a chance to take 0.992