Question

Suppose that in an assortment of 30 calculators there are 5 with defective switches. Draw with and without replacement. (Enter the probabilities as fractions.)

(a) If one machine is selected at random, what is the probability it has a defective switch?

with replacement

without replacement

(b) If two machines are selected at random, what is the probability that both have defective switches?

with replacement

without replacement

(c) If three machines are selected at random, what is the probability that all three have defective switches?

with replacement

without replacement

Answer 1

(a) If one machine is drawn(or selected) , then probaility that this would be defective is 5/30 or 1/6. There is no question of replacement or without replacement, because there is no further machine selection.

Prob of selecting defective macine = 1/6

(b) If two machines are selected, then probability of first machine being defective would be 1/6. If the machine taken out is replaced, that is, number of total machines remains 30, then probability of second machine selected is also defective would again be 1/6. Hence the probability of both the machines being defective would be 1/6 x 1/6 = 1/36

In the next case, that machine selected first is not replaced, the prbability would be as explained below:

The probability of first machine being defective would be 1/6. For selecting the second machine, there would be only 29 machines out which only 4 would be defective, as one defective machine has already gone out. The probability of selecting a defective second machine would thus be 4/29. The combined probability of selecting both defective machines would would thus be 1/6 x 4/29 = 2/87

Thus prob with replacement = 1/36

Prob without replacement = 2/87

c) Now if three machines are selected one by one with replacement, the probability of each of them being defective would be 1/6 and thus the combined probability would be 1/6 x 1/6 x 1/6 = 1/216

If each machine that is selected is not replaced, the probability of selecting first defective machine would be 1/6, prob of selecting second defective machine would be 4/29 and prob of selecting third defective machine would be 3/28 . The combined probability would be 1/6 x 4/29 x 3/28 = 1/406.

Thus prob with replacement = 1/216

Prob without replacment = 1/406

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