Chapter 4, Problem 7RE

a.

To determine

List all the outcomes in the sample space.

Answer to Problem 7RE

The sample space for the experiment is given by: $\left\{\text{DDD,DDG,DGD,GDD,DGG,GDG,GGD,GGG}\right\}$

Explanation of Solution

It is given that three circuits were installed in a computer. There were 12% chances that the circuit is defective. A defective circuit is denoted by ‘D’ and a good circuit by ‘G’.

Sample space:

The set of all possible outcomes of a probability experiment is called sample space of the experiment.

Here, each circuit has two possibilities D and G. Therefore, the sample space of the experiment is $\left\{\text{DDD,DDG,DGD,GDD,DGG,GDG,GGD,GGG}\right\}$. First, second and third letter in each outcome represents whether or not the first, second and third circuit is good or defective.

b.

To determine

Compute the probability that all three circuits are good.

Answer to Problem 7RE

The probability that all three circuits are good is 0.6815.

Explanation of Solution

Calculation:

Here, circuits are independents to each other.

Multiplication Rule for Independent Events:

For any two independent events A and B, $P\left(A\text{and }B\right)=P\left(A\right)\times P\left(B\right)$.

Required probability can be obtained as follows:

$P\left(\text{all three circuits are good}\right)=P\left(\text{first is good}\right)\times P\left(\text{second is good}\right)\times P\left(\text{third is good}\right)$

There are 12% chances that the circuit is defective. That is, $P\left(\text{Defective}\right)=0.12$.

Therefore,

$\begin{array}{c}P\left(\text{G}\right)=1-0.12\\ =0.88\end{array}$

Substitute this value in the above formula.

Therefore,

$\begin{array}{c}P\left(\text{all three circuits are good}\right)=P\left(\text{GGG}\right)\\ =\left(0.88\right)\times \left(0.88\right)\times \left(0.88\right)\\ ={\left(0.88\right)}^3\\ =0.68147\\ \approx 0.6815\end{array}$

Thus, the probability that all three circuits are good is 0.6815.

c.

To determine

Compute the probability that a computer will function.

Answer to Problem 7RE

The probability that a computer will function is 0.9603.

Explanation of Solution

Calculation:

The computer will function only when two or three of the circuits are good.

Required probability can be obtained as follows:

$P\left(\text{ two or three circuits are good}\right)=P\left(\text{GGD}\right)+P\left(\text{GDG}\right)+P\left(\text{DGG}\right)+P\left(\text{GGG}\right)$

From part (a), $P\left(\text{D}\right)=0.12\text{and}P\left(\text{G}\right)=0.88$. Substitute this value in the above formula and use the multiplication rule.

Therefore,

$\begin{array}{c}P\left(\text{ two or three circuits are good}\right)=P\left(\text{GGD}\right)+P\left(\text{GDG}\right)+P\left(\text{DGG}\right)+P\left(\text{GGG}\right)\\ =\left[\begin{array}{l}P\left(\text{G}\right)\times P\left(\text{G}\right)\times P\left(\text{D}\right)+P\left(\text{G}\right)\times P\left(\text{D}\right)\times P\left(\text{G}\right)+\\ P\left(\text{D}\right)\times P\left(\text{G}\right)\times P\left(\text{G}\right)+P\left(\text{G}\right)\times P\left(\text{G}\right)\times P\left(\text{G}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.88\right)\times \left(0.88\right)\times \left(0.12\right)+\left(0.88\right)\times \left(0.12\right)\times \left(0.88\right)+\\ \left(0.12\right)\times \left(0.88\right)\times \left(0.88\right)+\left(0.88\right)\times \left(0.88\right)\times \left(0.88\right)\end{array}\right]\\ =\left[3\times {\left(0.88\right)}^2\times \left(0.12\right)+{\left(0.88\right)}^3\right]\\ =0.278784+0.681472\\ =0.960256\\ =0.9603\end{array}$

Thus, the probability that computer will function is 0.9603.

d.

To determine

Check whether all three circuits to be defective are unusual or not.

Answer to Problem 7RE

All three circuits to be defective are unusual.

Explanation of Solution

Calculation:

The cut off is 0.05.

Unusual events:

An event is said to be unusual if the probability of occurrence of that event is less than 0.05.

Required probability can be obtained as follows:

$P\left(\begin{array}{l}\text{all three circuits }\\ \text{are defective}\end{array}\right)=\left[\begin{array}{l}P\left(\text{first is defective}\right)\times P\left(\text{second is defective}\right)\times \\ P\left(\text{third is defective}\right)\end{array}\right]$

There are 12% chances that the circuit is defective.  That is, $P\left(\text{D}\right)=0.12$ Substitute this value in the above formula.

Therefore,

$\begin{array}{c}P\left(\text{all three circuits are defective}\right)=P\left(\text{DDD}\right)\\ =\left(0.12\right)\times \left(0.12\right)\times \left(0.12\right)\\ ={\left(0.12\right)}^3\\ =0.001728\\ =0.0017\end{array}$

Thus, the probability that all three circuits are defective is 0.0017.

Here,

$\begin{array}{c}P\left(\text{all three circuits are defective}\right)=0.0017\\ <0.05\end{array}$

Therefore, event all three circuits to be defective are unusual.