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

a. 0.1024

b. 0.8976

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)=(n!)/(x!(n-x)!)*p^(x)*q^(n-x)\\P(x)=(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)=(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)=(2!)/(1!(2-1)!)*0.68^(1)*0.32^(2-1)=0.4352\\P(2)=(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.