Question

In a batch of 100 cell phones, there are, on average, 5 defective ones. if a random sample of 30 is selected, find the probability of having 2 defective phones.

Answer 1

Let x be the number of defective cell phones. It is given that in batch of 100, on average 5 are defective.

let p be the probability of defective cell phone.

p = 5/100 = 0.05

Let n be size of random sample, n=30

Here out of 30 we want to find probability that 2 will be defective. It means 30-2 =28 cell phones will be non defective.

The probability of getting non defective cell phone is 1- p=1-0.05 =0.95

The probabability of getting 2 defective is

P(X=2) = number of ways selecting 2 from 30 * probability 2 defective * probability of 28 non defective

Now number of ways of selecting 2 cell phone from 30 is

30C2 = $\frac{30!}{(30-2)! 2!}$

= $\frac{30!}{28! 2!}$

= $\frac{30 *29* 28!}{28! 2!}$

= (30*29) /2

30C2 = 435

P(X=2) = 30C2 * (0.05)^2 * (0.95)^28

= 435 * 0.0025 * 0.2378

P(X=2) = 0.2586

Probability of getting 2 defective out of 30 is 0.2586