An electronic product contains 40 integrated circuits. the probability that any integrated circuit is defective is 0.01, and the
integrated circuits are independent. the product operates only if there are no defective integrated circuits. what is the probability that the product operates?
We are given that the
operation of all circuits is independent with each other, therefore we can use
the multiplication rule for independent events, which states that P (intersection
of A and B) = P(A) * P(B). In this case, we want the intersection of circuit 1 to
be working with the intersection of circuit 2 on and on until circuit 40. That
is, we want every circuit to work with each other. The given probability that
circuit 1 works is .99. The probability that circuit 2 works is still .99 since
this is independent events. And we see that the probability for each of the 40
circuit to work is .99. <span>
So P (intersection of 1 through 40) = .99 * .99 *
.99.....*.99 = (.99)^40 = .6689717586</span>
Answer:
<span>There is a 0.67 probability
(or 67%) that the product will work.</span>
Okay, so if with 10 games, they win 6, they would win 12 with 20. To find the 5, you would find half of 6 (since half of 10 is 5). Half of 6 is three. 3 plus 12 is 15. They would win 15 games.
