Chapter 2.5, Problem 84E

a.

To determine

Find the probability that all the three components functions properly.

Answer to Problem 84E

The probability that all the three components functions properly is 0.7448.

Explanation of Solution

Given info:

The information is based on purchasing the audio components like a receiver, pair of speakers and a CD player. Here A₁ be the event of functioning of the receiver, A₂ be the event of functioning of the speaker  and A₃ be the event of functioning of CD player and its probabilities are $P\left(A_1\right)=0.95$, $P\left(A_2\right)=0.98$ and $P\left(A_3\right)=0.80$. Also, the events are independent.

Calculation:

Define the event as given below:

$\begin{array}{l}{A}_1\text{: the event of functioning of the receiver,}\\ A_2\text{: the event of functioning of the speaker and}\\ A_3\text{: the event of functioning of CD player}\end{array}$

Mutually independent events:

Two or more events are said to be mutually independent, if the occurrence of one does not affect the occurrence of other. Events $A_1,A_{2,}\mathrm{...}{A}_n$ is said to be mutually independent if for every k$\left(=2,3,\mathrm{...}n\right)$ and every subset $i_1,i_{2,}\mathrm{...}{i}_k$.

$P\left(A_{i_1}\cap A_{i_2}\cap \mathrm{...}\cap A_{i_k}\right)=P\left(A_{i_1}\right)\times P\left(A_{i_2}\right)\times \mathrm{...}\times P\left(A_{i_k}\right)$

The probability that all the three components functions properly can be obtained as

$\begin{array}{c}P\left(\text{receiver}\cap \text{Speaker}\cap \text{CDplayer}\right)=P\left(A_1\cap A_2\cap A_3\right)\\ =P\left(A_1\right)\times P\left(A_2\right)\times P\left(A_3\right)\\ =\left(0.95\right)\times \left(0.98\right)\times \left(0.80\right)\\ =0.7448\end{array}$

Thus, the probability of an event $\left(A_1\cap A_2\cap A_3\right)$ is 0.7448.

b.

To determine

Find the probability that at least one of the three components needs service.

Answer to Problem 84E

The probability that at least one of the three components needs service is 0.2552.

Explanation of Solution

Calculation:

The probability that at least one of the three components needs service is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{receiver needs service }\cup \\ \text{Speaker needs service}\cup \\ \text{CDplayer needs service}\end{array}\right)=P\left(A_1{}^{\prime }\cup A_2{}^{\prime }\cup A_3{}^{\prime }\right)\\ =1-P\left(A_1\cap A_2\cap A_3\right)\end{array}$

From previous part (a), $P\left(A_1\cap A_2\cap A_3\right)=0.7448$

The probability of event $\left(A_1{}^{\prime }\cup A_2{}^{\prime }\cup A_3{}^{\prime }\right)$ is given below:

$\begin{array}{c}P\left(A_1{}^{\prime }\cup A_2{}^{\prime }\cup A_3{}^{\prime }\right)=1-0.7448\\ =0.2552\end{array}$

Thus, the probability of an event $\left(A_1{}^{\prime }\cup A_2{}^{\prime }\cup A_3{}^{\prime }\right)$ is 0.2552.

c.

To determine

Find the probability that all the three components need service.

Answer to Problem 84E

The probability that all the three components need service is 0.0002.

Explanation of Solution

Calculation:

The probability that at least one of the three components needs service is obtained as:

$\begin{array}{c}P\left(\begin{array}{l}\text{receiver needs service }\cap \\ \text{Speaker needs service}\cap \\ \text{CDplayer needs service}\end{array}\right)=P\left(A_1{}^{\prime }\cap A_2{}^{\prime }\cap A_3{}^{\prime }\right)\\ =P\left(A_1{}^{\prime }\right)\times P\left(A_2{}^{\prime }\right)\times P\left(A_3{}^{\prime }\right)\end{array}$

The probability of the event $\left(A_1{}^{\prime }\cap A_2{}^{\prime }\cap A_3{}^{\prime }\right)$ is given below:

$\begin{array}{c}P\left(A_1\cap A_2\cap A_3\right)=P\left(A_1{}^{\prime }\right)\times P\left(A_2{}^{\prime }\right)\times P\left(A_3{}^{\prime }\right)\\ =\left(0.05\right)\times \left(0.02\right)\times \left(0.20\right)\\ =0.0002\end{array}$

Thus, the probability of an event $\left(A_1{}^{\prime }\cap A_2{}^{\prime }\cap A_3{}^{\prime }\right)$ is 0.0002.

d.

To determine

Find the probability that only receiver needs service.

Answer to Problem 84E

The probability that only receiver needs service is 0.0392.

Explanation of Solution

Calculation:

The probability that only receiver needs to be serviced can be obtained as

$\begin{array}{c}P\left(\begin{array}{l}\text{receiver needs service}\cap \\ \text{Speaker}\cap \text{CDplayer}\end{array}\right)=P\left(A_1{}^{\prime }\cap A_2\cap A_3\right)\\ =P\left(A_1{}^{\prime }\right)\times P\left(A_2\right)\times P\left(A_3\right)\\ =\left(0.05\right)\times \left(0.98\right)\times \left(0.80\right)\\ =0.0392\end{array}$

Thus, the probability that only receiver needs service is 0.0392.

e.

To determine

Find the probability that exactly one of the three components needs to be serviced.

Answer to Problem 84E

The probability that exactly one of the three components needs to be serviced is 0.2406.

Explanation of Solution

Calculation:

The probability that exactly one of the three components needs to be serviced is obtained as:

$\begin{array}{c}P\left(\left(A_1{}^{\prime }\cap A_2\cap A_3\right)\cup \left(A_1\cap A_2{}^{\prime }\cap A_3\right)\cup \left(A_1\cap A_2\cap A_3{}^{\prime }\right)\right)=\left[\begin{array}{l}P\left(A_1{}^{\prime }\cap A_2\cap A_3\right)+\\ P\left(A_1\cap A_2{}^{\prime }\cap A_3\right)+\\ P\left(A_1\cap A_2\cap A_3{}^{\prime }\right)\end{array}\right]\\ =\left[\begin{array}{l}P\left(A_1{}^{\prime }\right)\times P\left(A_2\right)\times P\left(A_3\right)+\\ P\left(A_1\right)\times P\left(A_2{}^{\prime }\right)\times P\left(A_3\right)+\\ P\left(A_1{}^{\prime }\right)\times P\left(A_2\right)\times P\left(A_3{}^{\prime }\right)\end{array}\right]\end{array}$

The probability of the event $\left(\left(A_1{}^{\prime }\cap A_2\cap A_3\right)\cup \left(A_1\cap A_2{}^{\prime }\cap A_3\right)\cup \left(A_1\cap A_2\cap A_3{}^{\prime }\right)\right)$ is given below:

$\begin{array}{c}P\left(\left(A_1{}^{\prime }\cap A_2\cap A_3\right)\cup \left(A_1\cap A_2{}^{\prime }\cap A_3\right)\cup \left(A_1\cap A_2\cap A_3{}^{\prime }\right)\right)=\left[\begin{array}{l}\left(0.05\right)\times \left(0.98\right)\times \left(0.80\right)+\\ \left(0.95\right)\times \left(0.02\right)\times \left(0.80\right)+\\ \left(0.95\right)\times \left(0.98\right)\times \left(0.20\right)\end{array}\right]\\ =0.0392+0.0152+0.1862\\ =0.2406\end{array}$

Thus, the probability of an event $\left(\left(A_1{}^{\prime }\cap A_2\cap A_3\right)\cup \left(A_1\cap A_2{}^{\prime }\cap A_3\right)\cup \left(A_1\cap A_2\cap A_3{}^{\prime }\right)\right)$ is 0.2406.

f.

To determine

Find the probability that all the three components functions properly that at least one of the components fails within a month after the warranty.

Answer to Problem 84E

The probability that the happening of the event that fails with a month after the warranty is zero.

Explanation of Solution

Justification:

There is no chance that the component fails with in the month after the warranty.

Thus, the probability for the occurrence of such event is zero.