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A quality control inspector has drawn a sample of 11 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that less than 7, but more than 4 bulbs from the sample are defective

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Answer:

0.04845

Step-by-step explanation:

Probability that less than 7 but more than 4 bulbs from the sample are defective :

P(4 < X < 7) = P(x = 5) + P(x = 6)

Using the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

From the question :

p = 0.2 ; 1 - p = 0.8 ; n = 11

P(x =5) = 11C5 * 0.2^5 * 0.8^6 = 0.03876

P(x = 6) = 11C6 * 0.2^6 * 0.8^5 = 0.00969

P(4 < X < 7) = P(x = 5) + P(x = 6)

P(4 < X < 7) = 0.03876 + 0.00969

P(4 < X < 7) = 0.04845

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