Chapter 11.4, Problem 19PPS

To determine

What is the probability that he makes exactly $7$ of his next $10$ kicks from within $35$ yards?

Answer to Problem 19PPS

  $\boxed{0.2503or25.03\%}$

Explanation of Solution

Given information:

A field goal kicker is accurate $75\%$ of the time from within $35$ yards. What is the probability that he makes exactly $7$ of his next $10$ kicks from within $35$ yards?

  $\begin{array}{cc}Range(yd)& Accuracy(\%)\\ 0-35& 75\\ 35-45& 62\\ 45+& 20\end{array}$

Calculation:

The probability of $k$ successes in $n$ trials of a binomial experiment is,

  $P(k){=}^n{C}_k{(p)}^k{(q)}^{n-k}$

Here $p$ represents the probability of success and $q$ represents the probability of failure which is equal to,

  $q=1-p$

Since the field kicker is $75\%$ accurate for the kicks within $35$ yards, so,

  $\begin{array}{l}p=75\%\\ =0.75\end{array}$

The probability of failure is,

  $q=1-p$

  $=1-0.75$

  $=0.25$

Substitute $p=0.75,q=0.25,k=7,n=10$ in the above formula to find the probability that the field goal kicker makes exactly $7$ out of his $10$ kicks within $35$ yards,

  $\begin{array}{l}P(7){=}^{10}{C}_7{(}^{0.75}{(}^{0.25}\\ =120{(}^{0.75}{(}^{0.25}\\ =0.2503\\ =25.03\%\end{array}$

Hence, the probability that the field goal kicker makes exactly $7$ out of his $10$ kicks within $35$ yards is $\boxed{0.2503or25.03\%}$