Mathematics, 14.02.2020 01:22 ellamai16

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.

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Answer from: jayy1474

a) There is a 10.24% probability that both generators fail during a power outage.

b) There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

Step-by-step explanation:

For each emergency backup generator, there are only two possible outcomes. Either they work correctly, or they fail. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

Assume that emergency backup generators fail 32% of the times when they are needed. So they work correctly 100-32 = 68% of the time. So p = 0.68

There are two generators, so n = 2

a. Find the probability that both generators fail during a power outage

This is P(X = 0)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.68)^{0}.(0.32)^{2} = 0.1024$

There is a 10.24% probability that both generators fail during a power outage.

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.

Either there are no working generators, or there is at least one working generator. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1024 = 0.8976$

There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

To be high enough for the hospital, this probability should be at least of 99%.

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