Math word problem

Question

A particular system is composed of four machines: A, B, C, and D. The entire system will work as long as any one (or more than one) of the four machines works. (With back up).

P(A works) = 0.60

P(B works) = 0.55

P(C works) = 0.65

P(D works) = 0.62

Rounded to the fourth digit beyond the decimal point, P(The entire system works) =

Answer 1

Assuming they are all independent (which they should be), i.e., one failing does not affect the probability that another will fail, the probability that ALL of them will fail is P(A)•(P(B)•P(C)•P(D)

Since the probability that A will work is 0.60, then the probability it will fail is 0.40

Since the probability that B will work is 0.55, then the probability it will fail is 0.45

Since the probability that C will work is 0.65, then the probability it will fail is 0.35

Since the probability that D will work is 0.62, then the probability it will fail is 0.38

So the probability that ALL will fail is (0.40)(0.45)(0.35)(0.38) = 0.02394 (or 2.394%)

The probability that NOT ALL will fail is 1 - 0.02394 = 0.97606 (or 97.606%)