Chapter 17, Problem 25E

(a)

To determine

To find: the probability that researcher gets all six serves in.

Answer to Problem 25E

0.12

Explanation of Solution

Given:

  $\begin{array}{c}n=6\\ r=6\end{array}$

  $p=0.7$

Formula used:

  $P\left(X=r\right)=C{}_r{\left(p\right)}^n{\left(q\right)}^r$

Calculation:

Required probability

  $\begin{array}{c}P\left(X=6\right)=C{}_6{\left(0.7\right)}^6{\left(0.3\right)}^0\\ =0.12\end{array}$

There is 12% possibility that all the six serves are successful

(b)

To determine

To find: the probability that researcher gets exactly four serves in.

Answer to Problem 25E

0.32

Explanation of Solution

Given:

  $\begin{array}{c}n=6\\ r=4\\ p=0.7\end{array}$

Formula used:

  $P\left(X=r\right)=C{}_r{\left(p\right)}^n{\left(q\right)}^r$

Calculation:

Required probability

  $\begin{array}{c}P\left(X=4\right)=C{}_4{\left(0.7\right)}^4{\left(0.3\right)}^2\\ =0.32\end{array}$

Therefore, there is 32% possibility that four out of six serves are successful.

(c)

To determine

To find: the probability that researcher gets minimum four serves in.

Answer to Problem 25E

0.74

Explanation of Solution

Given:

  $\begin{array}{c}n=6\\ p=0.7\end{array}$

Formula used:

  $P\left(X=r\right)=C{}_r{\left(p\right)}^n{\left(q\right)}^r$

Calculation:

Required probability

  $\begin{array}{c}P\left(X\ge 4\right)=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ =C{}_4{\left(0.7\right)}^4{\left(0.3\right)}^2+C{}_5{\left(0.7\right)}^5\left(0.3\right)+C{}_6{\left(0.7\right)}^6{\left(0.3\right)}^0\\ =0.74\end{array}$

There is 74% possibility that minimum four servers are successful.

(d)

To determine

To find: the probability that researcher gets not greater than four serves in.

Answer to Problem 25E

0.58

Explanation of Solution

Given:

  $\begin{array}{c}n=6\\ p=0.7\end{array}$

Formula used:

  $P\left(X=r\right)=C{}_r{\left(p\right)}^n{\left(q\right)}^r$

Calculation:

Required probability

  $\begin{array}{c}P\left(X\le 4\right)={\displaystyle \sum _{x=0}^{4}C{}_x{\left(0.7\right)}^x{\left(0.3\right)}^{6-x}}\\ =0.58\end{array}$

There is 58% possibility that not greater than 4 serves are successful.