Question

1. a complex electronic device contains three components, a, b, and c. the probabilities of failure for each component in any one year are 0.01, 0.03, and 0.04, respectively. if any one component fails, the device will fail. if the components fail independently of one another, what is the probability that the device will not fail in one year? (a) less than 0.01 (b) 0.078 (c) 0.080 (d) 0.922 (e) greater than 0.99

Explanation:

Step1: Calculate the probability of each component not failing

The probability that component A does not fail in one year is \(P(\text{not }A)=1 - 0.01=0.99\).
The probability that component B does not fail in one year is \(P(\text{not }B)=1 - 0.03 = 0.97\).
The probability that component C does not fail in one year is \(P(\text{not }C)=1 - 0.04=0.96\).

Step2: Use the multiplication rule for independent events

Since the components fail independently, the probability that the device does not fail (i.e., all components do not fail) is \(P = P(\text{not }A)\times P(\text{not }B)\times P(\text{not }C)\).
Substitute the values: \(P=0.99\times0.97\times0.96\).
First, \(0.99\times0.97=(1 - 0.01)\times0.97=0.97-0.0097 = 0.9603\).
Then, \(0.9603\times0.96=(0.96 + 0.0003)\times0.96=0.96\times0.96+0.0003\times0.96\).
We know that \(0.96\times0.96=(1 - 0.04)\times(1 - 0.04)=1-0.08 + 0.0016=0.9216\), and \(0.0003\times0.96 = 0.000288\).
So \(P=0.9216+0.000288=0.921888\approx0.922\).

Answer:

D. 0.922