<PRE><B>Question 676296</B>
There are eight bulbs in a box, five of which are defective. If three bulbs are selected at random, without replacement, find the following probability:

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(a) All three are defective.
# of ways to succeed = 5C3 = 10
# of possible outcomes = 8C3 = 56
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P(x = 3) = 5C3/8C3 = 10/56 = 5/28
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(b)None are defective.
# of ways to succeed = 3C3 = 1
# of possible outcomes = 8C3 = 56
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P(x = 0) = 3C3/8C3 = 1/56

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Cheers,
Stan H.
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