Question

12. Suppose you receive a shipment of 33 computers, two of which are defective. What is the probability of picking a sample of 6 computers with exactly one bad computer in it? (Round approximations to the nearest thousandth.)

Answer 1

Oh-kay.
So, that means, we 're doin' two things here:

first, picking up one defective comp. out of the 33, where two are defective ->
so, no. of ways of selecting any 1 out of the two defective = 2 --> favorable outcomes.
no. of ways of selecting any 1 comp. out of 33 = 33
so, probab. of selecting the defective comp. = (2/33)

next, out of the the remaining 32 comps., out of which 31 are good, I choose 5 good ones.
So, favorable outcomes = 31-CHOOSE-5 = 31C5
total outcomes - choosing any 5 out of all 32 remaining = 32C5
so, probab. = (31C5) / (32C5)


Now, probability of both the events occurring simultaneously --> product of their respective probabilities = (2/33) * ((31C5) / (32C5))




helps?

Answer 2

@Curry okay now?