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For a sample of n=6 lightbulbs. The probability of a broken bulb is 9%. What is the probability of exactly two broken bulbs?
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Answer #1

Please note ^nC_r = \frac{n!}{(n-r)! * r!}

Binomial Probability = nCx * (p)x * (q)n-x, where n = number of trials and x is the number of successes.

Here n = 6, p = 0.09, q = 1 – p = 0.91.

P(X = 2) = 6C2 * (0.09)2 * (0.91)6-2 = 0.0833

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