Question

Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backup generators fail 32% of the times when they are needed. A hospital has two backup generators so that power is available if one of them fails during a power outage. Complete parts (a) and (b) below.
a. Find the probability that both generators fail during a power outage (Round to four decimal places as needed.)
b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital? Assume the hospital needs both generators to fail less than 1% of the time when needed. (Round to four decimal places as needed.)

Answer 1

Answer:

a. 0.1024

b. 0.8976

Step-by-step explanation:

The probability that x generators don't fail when they are needed follows a binomial distribution, because we have n identical and independent events (2 backup generators) with a probability p of success (1-0.32=0.68) and a probability q of failure (0.32).

So, the probability that x generator success are calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*q^{n-x}\\P(x)=\frac{2!}{x!(2-x)!}*0.68^{x}*0.32^{2-x}$

Then, the probability that both generators fail during a power outage is equal to the probability that 0 generators success. It is calculated as:

$P(0)=\frac{2!}{0!(2-0)!}*0.68^{0}*0.32^{2-0}=0.1024$

At the same way, the probability of having a working generator in the event of a power outage is equal to the probability that at least 1 generator success. It is calculated as:

$P(x\geq1)=P(1)+P(2) \\P(1)=\frac{2!}{1!(2-1)!}*0.68^{1}*0.32^{2-1}=0.4352\\P(2)=\frac{2!}{2!(2-2)!}*0.68^{2}*0.32^{2-2}=0.4624\\P(x\geq1)=0.4352+0.4624=0.8976$

This probability is not high enough for the hospital, both generators fail approximately the 10% of the time when needed.