SOLUTION: In a sample of 10 discs, it is known that 2 are defective. Three discs are selected at random and without replacement from this sample. What is the probability that: a) all three

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Question 274529: In a sample of 10 discs, it is known that 2 are defective. Three discs are selected at random and without replacement from this sample. What is the probability that:
a) all three are defective
b) none is defective
c) two only are defective

can you please give me the answer with the solution? thanks
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
probability that 0 out of 3 are defective is 8/10 * 7/9 * 6/8 * 1 = .466666667

probability that 1 out of 3 are defective is 2/10 * 8/9 * 7/8 * 3 = .466666667

probability that 2 out of 3 are defective is 2/10 * 1/10 * 8/8 * 3 = .066666666

sum of all probabilities equals 1.

probability that 3 out of 3 are defective is 2/10 * 1/10 * 0/8 = 0

the multipliers are as follows:

set of 3 with all defective = ddd = multiplier of 1.

set of 3 with 1 defective = dnn + ndn + nnd = multiplier of 3.

set of 3 with 2 defective = ddn + dnd + ndd = multiplier of 3.

to see how the multipliers work, consider the probability that 1 out of 3 is defective.

p(dnn) = 2/10 * 8/9 * 7/8 = (2*8*7)/(10*9*8) = 112/720 = .155555555

p(ndn) = 8/10 * 2/9 * 7/8 = (8*2*7)/(10*9*8) = 112/720 = .155555555

p(nnd) = 8/10 * 7/9 * 2/8 = (8*7*2)/(10*9*8) = 112/720 = .155555555

The probability of each of these occurrences is the same so you take one of them and multiply it by 3 and you get the same answer.