Chapter 6.5, Problem 51E

a.

To determine

Compute the probability of 1-2 subsystem works only if both components work.

Answer to Problem 51E

The probability of 1-2 subsystem works only if both components work is 0.81.

Explanation of Solution

Calculation:

The given system consists of four components and all of the four components work independently.

Multiplication Rule:

If two Events A and B are independent then $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$.

Then, the required probability is obtained by the following formula:

$P\left(\text{1-2 subsystem works}\right)=P\left(\text{1 work}\right)\times P\left(\text{2 work}\right)$

It is given that $P\left(1\text{ works}\right)=P\left(2\text{works}\right)=P\left(3\text{ works}\right)=P\left(4\text{ works}\right)=0.9$. Substitute these values in the above equation.

$\begin{array}{c}P\left(\text{1-2 subsystem works}\right)=0.9\times 0.9\\ =0.81\end{array}$

Thus, the probability of 1-2 subsystem works only if both components work is 0.81.

b.

To determine

Obtain the probability of 1-2 subsystem does not work.

Find the probability of 3-4 subsystem does not work.

Answer to Problem 51E

The probability of 1-2 subsystem does not work is 0.19.

The probability of 3-4 subsystem does not work is 0.19.

Explanation of Solution

Calculation:

From Part (a), the probability of 1-2 subsystem works only if both components work is 0.81.

Then, the probability of 1-2 subsystem does not work is obtained by the following formula:

$\begin{array}{c}P\left(\text{1-2 subsystem does not work}\right)=1-P\left(\text{1-2 subsystem work}\right)\\ =1-0.81\\ =0.19\end{array}$

Thus, the probability of 1-2 subsystem does not work is 0.19.

The probability of 3-4 subsystem works is obtained by the following formula:

$\begin{array}{c}P\left(\text{3-4 subsystem works}\right)=P\left(\text{3 work}\right)\times P\left(\text{4 work}\right)\\ =0.9\times 0.9\\ =0.81\end{array}$

Then, the probability of 3-4 subsystem does not work is obtained by the following formula:

$\begin{array}{c}P\left(\text{3-4 subsystem does not work}\right)=1-P\left(\text{3-4 subsystem work}\right)\\ =1-0.81\\ =0.19\end{array}$

Thus, the probability of 3-4 subsystem does not work is 0.19.

c.

To determine

Compute the probability that subsystem would not work if the 1-2 subsystem does not work and if 3-4 subsystem also does not work.

Compute the probability that subsystem will work.

Answer to Problem 51E

The probability that subsystem would not work is 0.0361.

The probability of subsystem will work is 0.9639.

Explanation of Solution

Calculation:

From Part (b), the probability of 1-2 subsystem does not work is 0.19 and the probability of 3-4 subsystem does not work is 0.19.

The probability of 3-4 subsystem works is obtained by the following formula:

$\begin{array}{c}P\left(\text{subsystem won't works}\right)=\left(\begin{array}{c}P\left(\text{1-2 subsystem does not work}\right)\times \\ P\left(\text{3-4 subsystem does not work}\right)\end{array}\right)\\ =0.19\times 0.19\\ =0.0361\end{array}$

Thus, the probability that subsystem would not work is 0.0361.

Then, the probability of subsystem will work is obtained by the following formula:

$\begin{array}{c}P\left(\text{subsystem will work}\right)=1-0.0361\\ =0.9639\end{array}$

Thus, the probability of subsystem will work is 0.9639.

d.

To determine

Explain how would be the probability of the system working change if a 5-6 subsystem were added in parallel with the other two subsystems.

Explanation of Solution

Calculation:

The given system consists of four components and all of the four components work independently.

Multiplication Rule:

If two Events A and B are independent then $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$.

From Part (b), the probability of 1-2 subsystem does not work is 0.19 and the probability of 3-4 subsystem does not work is 0.19.

It is given that 5-6 subsystem is also added in parallel, then the probability of 5-6 subsystem works only if both components work is 0.81.

Then, the probability of 5-6 subsystem does not work is obtained by the following formula:

$\begin{array}{c}P\left(\text{5-6 subsystem does not work}\right)=1-P\left(\text{5-6 subsystem work}\right)\\ =1-0.81\\ =0.19\end{array}$

Thus, the probability of 5-6 subsystem does not work is 0.19.

The probability of subsystem would not work is obtained by the following formula:

$\begin{array}{c}P\left(\text{subsystem won't works}\right)=\left(\begin{array}{c}P\left(\text{1-2 subsystem does not work}\right)\times \\ P\left(\text{3-4 subsystem does not work}\right)\times \\ P\left(\text{5-6 subsystem does not work}\right)\end{array}\right)\\ =0.19\times 0.19\times 0.19\\ =0.006859\end{array}$

Thus, the probability that subsystem would not work is 0.006859.

Then, the probability of subsystem will work is obtained by the following formula:

$\begin{array}{c}P\left(\text{subsystem will work}\right)=1-0.006859\\ =0.993141\end{array}$

Thus, the probability of subsystem will work is 0.993141.

e.

To determine

Describe how would be the probability that the system works change if there were three components in series in each of the two subsystems.

Explanation of Solution

Calculation:

The probability that one particular subsystem will work is obtained by the following formula:

$\begin{array}{c}P\left(\text{one particular subsystem will work}\right)=\left(0.9\right)\left(0.9\right)\left(0.9\right)\\ =0.729\end{array}$

Thus, the probability that one particular subsystem will work is 0.729.

Then, the probability that the subsystem will not work is as follows:

$\begin{array}{c}P\left(\text{subsystem will not works}\right)=1-0.729\\ =0.271\end{array}$

Then, the probability that the subsystem will not work is 0.271.

Then, the probability that neither of the two subsystems work is as follows:

$\begin{array}{c}P\left(\text{system does not work}\right)=\left(0.271\right)\left(0.271\right)\\ =0.073441\end{array}$

Thus, the probability that neither of the two subsystems work is 0.073441.

Therefore, the probability that the system works is as follows:

$\begin{array}{c}P\left(\text{system works}\right)=1-0.073441\\ =0.926559\end{array}$

Thus, the probability that the system works is 0.926559.