Question

A boiler has five identical relief valves. the probability that any particular valve will open on demand is .95. assuming independent operation of the valves, calculate p(at least one valve opens) and p(at least one valve fails to open).

Answer 1

To calculate P(at least one valve opens), we can find 1 - p(no valves open).
This is equal to 1 - (0.05)^5, or 0.9999996875. 
P(at least one valve fails to open) is equal to 1 - p(all valves open and none fail). This is 1 - (0.95)^5, or 0.2262190625.

Answer 2

Answer:

p(at least one valve opens) = 0.9999996875

p(at least one valve fails to open) = 0.2262

Step-by-step explanation:

For each valve, there are only two possible outcomes. Either it opens in demand, or it does not. The probability of a valve opening on demand is independent of other valves. So we use the binomial probability distribution to solve this question.

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.

A boiler has five identical relief valves.

This means that $n = 5$

The probability that any particular valve will open on demand is .95

This means that $p = 0.95$

p(at least one valve opens)

Either no valve opens, or at least one does. 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

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

In which

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

$P(X = 0) = C_{5,0}.(0.95)^{0}.(0.05)^{5} = 0.0000003125$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000003125 = 0.9999996875$

p(at least one valve opens) = 0.9999996875

p(at least one valve fails to open).

Either all five valves open, or les than five do. If less than five open, it means that at least one failed to open. So

$P(X = 5) + P(X < 5) = 1$

We want P(X < 5). So

$P(X < 5) = 1 - P(X = 5)$

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

$P(X = 5) = C_{5,5}.(0.95)^{5}.(0.05)^{0} = 0.7738$

$P(X < 5) = 1 - P(X = 5) = 1 - 0.7738 = 0.2262$

p(at least one valve fails to open) = 0.2262