Chapter 9.7, Problem 47E

a.

To determine

To find: the probability that the all four units are good.

Answer to Problem 47E

  $\frac{14}{55}$

Explanation of Solution

Given:

Total number of ovens = 12

Total number of defective oven = 3

Total number of ovens that are good = $12-3=9$

Total number of ovens purchased = 4

Let E be the event that all four units are good.

The possible number of ways to choose 4 out of the 9 good ovens is:

  $\begin{array}{c}n\left(E\right)=C{}_4\\ =\frac{\text{9!}}{\text{4!}\left(9-4\right)\text{!}}\\ =126\end{array}$

Total possible number of ways to choose 4 out of 12 ovens is:

  $\begin{array}{c}n\left(S\right)=C{}_4\\ =\frac{\text{12!}}{\text{4!}\left(12-4\right)\text{!}}\\ =495\end{array}$

Thus, the probability that all four units are good is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{126}{495}\\ =\frac{14}{55}\end{array}$

b.

To determine

To find: the probability that exactly two units are good.

Answer to Problem 47E

  $\frac{12}{55}$

Explanation of Solution

Given:

Total number of ovens = 12

Total number of defective oven = 3

Total number of ovens that are good = $12-3=9$

Total number of ovens purchased = 4

Let E be the event that exactly two units are good.

That is, 2 ovens are good and 2 are defective. So,

  $\begin{array}{c}n\left(E\right)=C{}_2\times C{}_2\\ =\frac{\text{9!}}{\text{2!}\left(9-2\right)\text{!}}\times \frac{\text{3!}}{\text{2!}\left(3-2\right)\text{!}}\\ =36\times 3\\ =108\end{array}$

Total possible number of ways to choose 4 out of 12 ovens is:

  $\begin{array}{c}n\left(S\right)=C{}_4\\ =\frac{\text{12!}}{\text{4!}\left(12-4\right)\text{!}}\\ =495\end{array}$

Thus, the probability that exactly two units are good is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{108}{495}\\ =\frac{12}{55}\end{array}$

c.

To determine

To find: the probability that at least two units are good.

Answer to Problem 47E

  $\frac{54}{55}$

Explanation of Solution

Given:

Total number of ovens = 12

Total number of defective oven = 3

Total number of ovens that are good = $12-3=9$

Total number of ovens purchased = 4

Let E be the event that at least two units are good.

(It is the sum of the cases exactly 2 units are good, exactly 3 units are good and all 4 units are good)

  $\begin{array}{c}n\left(E\right)=C{}_2\times C{}_2+C{}_3\times C{}_1+C{}_4\\ =\frac{\text{9!}}{\text{2!}\left(9-2\right)\text{!}}\times \frac{\text{3!}}{\text{2!}\left(3-2\right)\text{!}}+\frac{\text{9!}}{\text{3!}\left(9-3\right)\text{!}}\times \frac{\text{3!}}{\text{1!}\left(3-1\right)\text{!}}+\frac{\text{9!}}{\text{4!}\left(9-4\right)\text{!}}\\ =36\times 3+84\times 3+126\\ =486\end{array}$

Total possible number of ways to choose 4 out of 12 ovens is:

  $\begin{array}{c}n\left(S\right)=C{}_4\\ =\frac{\text{12!}}{\text{4!}\left(12-4\right)\text{!}}\\ =495\end{array}$

Thus, the probability that at least two units are good is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{486}{495}\\ =\frac{54}{55}\end{array}$