From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
(1) 73108
With replacement, the 8 draws are Bernoulli trials with p = 10100 = 110, q = 910.
P(X 7) = P(X = 7) + P(X = 8)
= 8C7 ( 110 )7 ( 910 ) + ( 110 )8
= 8 9108 + 1108 = 73108
Answer: (1)
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