A quality control inspector has drawn a sample of 10 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that exactly 7 bulbs from the sample are defective? Round your answer to four decimal places

Answer:The probability is 0.0008.Step-by-step explanation:Let X represents the event of defective bulb,Given, the probability of defective bulb, p = 20 % = 0.2,So, the probability that bulb is not defective, q = 1 - p = 0.8,The number of bulbs drawn, n = 10,Since, binomial distribution formula,[tex]P(x=r) = ^nC_r p^r q^{n-r}[/tex]Where, [tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex]Hence, the probability that exactly 7 bulbs from the sample are defective is,[tex]P(X=7)=^{10}C_7 (0.2)^7 (0.8)^{10-7}[/tex][tex]=120 (0.2)^7 (0.8)^3[/tex][tex]=0.000786432[/tex][tex]\approx 0.0008[/tex]