Question

2.31 A boxcar contains six complex electronic systems. Two of the six are to be randomly selectec for thorough testing and then classified as defective or not defective. a If two of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective. If four of the six systems are actually defective, find the probabilities indicated in part (a) b

Answer 1

a. The probability of at least one is the same as one minus the probability of none.

Let C(a,b) denotes be a choose be or choosing b things from a, then the probability that none of them is defective is:

[C(4,2)*C(2,0)]/C(6,2) = 6/15

1-6/15 = 9/15

so the probability of at least one of the two systems are defective is 9/15.

The probability of both are defective:

[C(2,2)*C(4,0)]/C(6,2) = 1/15

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b. Using the same reasoning as above, we have

i) 1- [C(2,2)*C(4,0)]/C(6,2) = 1- 1/15 = 14/15

ii) [C(4,2)*C(2,0)]/C(6,2) = 6/15