Chapter 2, Problem 1SE

A system consists of four components connected as shown.

Assume A, B, C, and D function independently. If the probabilities that A, B, C, and D fail are 0.1, 0.2, 0.05, and 0.3, respectively, what is the probability that the system functions?

To determine

Find the probability that the system functions.

Answer to Problem 1SE

The probability that the system functions is 0.9997.

Explanation of Solution

Given info:

A, B, C, and D are functions independently. The probabilities that A, B, C, and D fail are 0.1, 0.2, 0.05, and 0.3, respectively.

Calculation:

Let A be the event that component A functions, let B be the event that component B functions, let C be the event that component C functions, and let D be the event that component D functions.

Therefore, the probability of A is,

$\begin{array}{c}P\left(A\right)=1-0.1\\ =0.9\end{array}$

The probability of B is,

$\begin{array}{c}P\left(B\right)=1-0.2\\ =0.8\end{array}$

The probability of C is,

$\begin{array}{c}P\left(C\right)=1-0.05\\ =0.95\end{array}$

The probability of D is,

$\begin{array}{c}P\left(D\right)=1-0.3\\ =0.7\end{array}$

From the given figure, it is observed that the system functions is,

$\left(A\cup B\right)\cup \left(C\cup D\right)$.

Therefore,

$P\left(\left(A\cup B\right)\cup \left(C\cup D\right)\right)=P\left(A\cup B\right)+P\left(C\cup D\right)-P\left(A\cup B\right)P\left(C\cup D\right)$

The probability of $\left(A\cup B\right)$ is obtained as follows:

$\begin{array}{c}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)\\ =0.9+0.8-\left(0.9\right)\left(0.8\right)\\ =1.7-0.72\\ =0.98\end{array}$

The probability of $\left(C\cup D\right)$ is obtained as follows:

$\begin{array}{c}P\left(C\cup D\right)=P\left(C\right)+P\left(D\right)-P\left(C\right)P\left(D\right)\\ =0.95+0.7-\left(0.95\right)\left(0.7\right)\\ =1.65-0.665\\ =0.985\end{array}$

Therefore, $P\left(\left(A\cup B\right)\cup \left(C\cup D\right)\right)$ becomes,

$\begin{array}{c}P\left(\left(A\cup B\right)\cup \left(C\cup D\right)\right)=0.98+0.985-\left(0.98\right)\left(0.985\right)\\ =1.965-0.9653\\ =0.9997\end{array}$

Hence, the probability that the system functions is 0.9997.