Chapter 11.6, Problem 33PPS

(a)

To determine

Find the likelihood that he makes 9 of his next 10 attempts from within 40 yards.

Answer to Problem 33PPS

The likelihood that he makes 9 of his next 10 attempts from within 40 yards is 0.121

Explanation of Solution

Given:

In $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$ , let p represent the likelihood of a success and q represent the likelihood of a failure.

If a place − kicker makes 70% of his kicks within 40 yards, find the likelihood that he makes 9 of his next 10 attempts from within 40 yards.

Concept Used:

Substitute $n=10,k=9,p=0.7\text{and}q=0.3$ in $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$

Calculation:

  $\begin{array}{l}n=10,k=9,p=0.7\text{and}q=0.3\\ $ \frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}=\frac{10!}{10!(10-9)!}{(0.7)}^9(0.3)$$ =\frac{10!}{10!1!}{(0.7)}^9(0.3)$$ =10{(0.7)}^9(0.3)=0.121$\end{array}$

Thus, the likelihood that he makes 9 of his next 10 attempts from within 40 yards is 0.121

(b)

To determine

Find the likelihood that he completes 8 of his next 10 attempts.

Answer to Problem 33PPS

The likelihood that he completes 8 of his next 10 attempts is 0.121

Explanation of Solution

Given:

In $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$ , let p represent the likelihood of a success and q represent the likelihood of a failure.

If a quarterback completes 60% of his passes, find the likelihood that he completes 8 of his next 10 attempts.

Concept Used:

Substitute: $n=10,k=8,p=0.6\text{and}q=0.4$ in $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$

Calculation:

  $\begin{array}{l}n=10,k=8,p=0.6\text{and}q=0.4\\ $ \frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}=\frac{10!}{8!(10-8)!}{(0.6)}^8{(0.4)}^2$$ =\frac{10!}{8!2!}{(0.6)}^8{(0.4)}^2$$ =45{(0.6)}^8{(0.4)}^2=0.121$\end{array}$

Thus,the likelihood that he completes 8 of his next 10 attempts is 0.121

(c)

To determine

Find the likelihood that they converts 2 of their next 5 conservations.

Answer to Problem 33PPS

The likelihood that they converts 2 of their next 5 conservations is 0.309.

Explanation of Solution

Given:

In $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$ , let p represent the likelihood of a success and q represent the likelihood of a failure.

If a team converts 30% of their two − point conversations, find the likelihood that they converts 2 of their next 5 conservations.

Concept Used:

Substitute: $n=5,k=2,p=0.3\text{and}q=0.7$ in $\frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}$

Calculation:

  $\begin{array}{l}n=5,k=2,p=0.3\text{and}q=0.7\\ $ \frac{n!}{k!(n-k)!}{p}^k{q}^{n-k}=\frac{5!}{2!(5-2)!}{(0.3)}^2{(0.7)}^3$$ =\frac{5!}{2!3!}{(0.3)}^2{(0.7)}^3$$ =10{(0.3)}^2{(0.7)}^3=0.309$\end{array}$

Thus, the likelihood that they converts 2 of their next 5 conservations is 0.309.