Question

Suppose 30% of items produced in a factory are defective and 12 items are chosen at random. Find the probability that exactly 4 of the 12 selected items are defective.

Answer 1

The probability that exactly 4 of the 12 selected items are defective is 0.2311.

Step-by-step explanation:

We are given that 30% of items produced in a factory are defective and 12 items are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 12 items

            r = number of success = exactly 4

           p = probability of success which in our question is % of items

                  produced in a factory that are defective, i.e; 30%

LET X = Number of defective items

So, X ~ Binom(n = 12, p = 0.30)

Now, probability that exactly 4 of the 12 selected items are defective is given by = P(X = 4)

                 P(X = 4)  =  $\binom{12}{4}\times 0.30^{4}\times (1-0.30)^{12-4}$

                                =  $495\times 0.30^{4}\times 0.70^{8}$

                                =  0.2311

Therefore, the probability that exactly 4 of the 12 selected items are defective is 0.2311.