Question

A TV set contains five circuit boards of type A, five of type B, and four of type C. The probability of failing in its first 5000 hours of use is 0.03 for a type A circuit board, 0.04 for a type B circuit board, and 0.03 for a type C circuit board. Assuming that the failures of the various circuit boards are independent of one another, compute the probability that no circuit board fails in the first 5000 hours of use.

Answer 1

Answer:

0.903264

Step-by-step explanation:

Given that a TV set contains five circuit boards of type A, five of type B, and four of type C. The probability of failing in its first 5000 hours of use is 0.03 for a type A circuit board, 0.04 for a type B circuit board, and 0.03 for a type C circuit board.

Let A' = the event that A fails, B' =  B fails and C'= C fails.

Probability that no circuit board fails = Prob (A'B'C')

= P(A')P(B')P(C')  (since A,B, C are independent, their complements are independent

= (1-0.03)(1-0.04)(1-0.03)

= 0.903264