Chapter 8.6, Problem 63E

(a)

To determine

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

Answer to Problem 63E

The probability that all four units would be good is $\frac{14}{55}$ .

Explanation of Solution

Given information:

A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random.

Calculation:

Let us try to solve the problem by first finding out many units are perfect

Out of 12 ovens, 3 are defective, thus the number of perfect units are $12-3=9$ units

After the first good unit is selected, the perfect units left would be 8 and total number of units would be 11

After the second good unit is selected, the perfect units left would be 7 and total number of units would be 10

After the third good unit is selected, the perfect units left would be 6 and total number of units would be 9

Thus the probability that all four units would be good

  $\begin{array}{l}=\frac{9}{12}\times \frac{8}{11}\times \frac{7}{10}\times \frac{6}{9}\\ =\frac{14}{55}\end{array}$

Thus, the probability that all four units would be good is $\frac{14}{55}$ .

(b)

To determine

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

Answer to Problem 63E

The probability that exactly two units would be good is $\frac{12}{55}$

Explanation of Solution

Given information:

A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random.

Calculation:

Let us try to find the probability that exactly 2 units are good.

Out of 12 ovens, 3 are defective, thus the number of perfect units are $12-3=9$ units

After the first good unit is selected, the perfect units left would be 8 and total number of units would be 11

After the second good unit is selected, the perfect units left would be 7 and total number of units would be 10

After the first bad unit is chosen, bad units left would be 2 and total number of units would be 9

Thus the probability that exactly two units would be good is

  $\begin{array}{l}=\frac{9}{12}\times \frac{8}{11}\times \frac{3}{10}\times \frac{2}{9}{\times }^4{C}_2\\ =\frac{2}{55}\times \frac{4!}{2!\times 2!}\\ =\frac{12}{55}\end{array}$

Thus, the probability that exactly two units would be good is $\frac{12}{55}$

(c)

To determine

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

Answer to Problem 63E

The probability that eat least two units would be good is $\frac{54}{55}$ .

Explanation of Solution

Given information:

A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random.

Calculation:

Let us try to find the probability that at least two units good.

Out of 12 ovens, 3 are defective, thus the number of perfect units are $12-3=9$ units

After the first good unit is selected, the perfect units left would be 8 and total number of units would be 11

After the second good unit is selected, the perfect units left would be 7 and total number of units would be 10

Out of 12 ovens, 3 are defective

Thus the probability that, three units would be good is

  $\begin{array}{l}=\frac{{}^9{C}_2{\times }^3{C}_2{+}^9{C}_3{\times }^3{C}_1{+}^9{C}_4}{{}^{12}{C}_4}\\ =\frac{108+252+126}{495}\\ =\frac{486}{495}\\ =\frac{54}{55}\end{array}$

Thus, the probability that eat least two units would be good is $\frac{54}{55}$ .