Question

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 1

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,

$P(x=r) = ^nC_r p^r q^{n-r}$

Where, $^nC_r = \frac{n!}{r!(n-r)!}$

Hence, the probability that exactly 7 bulbs from the sample are defective is,

$P(X=7)=^{10}C_7 (0.2)^7 (0.8)^{10-7}$

$=120 (0.2)^7 (0.8)^3$

$=0.000786432$

$\approx 0.0008$