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

A: Event of valve open on demand

P(A) = 0.95

B : Event of valve fail to open

P(B) = 1- P(A) = 1-0.95 = 0.05

Probability of atleast one valve opens = 1 - Probability of all 5 valves fail to open = 1 - P(B $\small \cap$ B $\small \cap$ B $\small \cap$ B $\small \cap$ B)

= 1- (P(B) x P(B) xP(B) xP(B) xP(B) )

= 1 - (0.05x0.05x0.05x0.05x0.05)

=1 - (0.0000003125) = 0.9999996875

Probability of atleast one valve fails to opens = 1 - Probability of all 5 valves open = 1 - P(A $\small \cap$ A $\small \cap$ A $\small \cap$ A $\small \cap$ A)

= 1- (P(A) x P(A) xP(A) xP(A) xP(A) )

= 1 - (0.95x0.95x0.95x0.95x0.95)

= 1- 0.7737809375

=0.2262190625