College Algebra, Chapter 10, 10.4, Section 10.4, Problem 36

An archer normally hits the target with probability of $0.6$. She hires a new coach for a series of special lessons. After the lessons, she hits the target in five out of eight attempts.

a.) Find the probability that she would have hit five or more out of the eight attempts before her lessons with the new coach.

b.) Did the new coaching appear to make a difference? (Consider the coaching effective if the probability in part (a) is $0.05$ or less.)

Recall that the formula for the binomial probability is given by

$C(n,r) p^r q^{n-r}$

a.) In this case, where the archer has not hire a coach yet, the probability of success is $p=0.60$ while the probability of failure is $q=1-p=0.40$. Thus, the probability that the archer hits the target five or more eight attempts is equal to sum of the probability that the archer hits exactly $5,6,7$ and $8$ of her attempts. Thus, we have


$
\begin{equation}
\begin{aligned}

=& C(8,5) (0.60)^5 (0.40)^{8-5} +C(8,6) (0.60)^6 (0.40)^{8-6} + C(8,7) (0.60)^7 (0.40)^{8-7} + C(8,8) (0.60)^8 (0.40)^{8-8}
\\
\\
=& 0.2787 + 0.2090 + 0.0896 + 0.0168
\\
\\
=& 0.5491

\end{aligned}
\end{equation}
$



b.) After hiring a new coach, the probability of success is $\displaystyle p= \frac{5}{8} = 0.625$. This gives the probability of failure $q=1-p=0.375$. Thus, the probability in this case is


$
\begin{equation}
\begin{aligned}

=& C(8,5) (0.625)^5 (0.375)^{8-5} +C(8,6) (0.625)^6 (0.375)^{8-6} + C(8,7) (0.625)^7 (0.375)^{8-7} + C(8,8) (0.625)^8 (0.375)^{8-8}
\\
\\
=& 0.2816+0.2347+0.1118+0.0233
\\
\\
=& 0.6514


\end{aligned}
\end{equation}
$



It shows that the coaching is effective since probability that the archer hits her target in the given case is increased by $0.6514 - 0.5941 = 0.0573$.