Question

a box contains 30 mobile s 6of which are defective mobile the probability that if a sample of3 mobile are chosen at random from the box 2 will be defective mobile

Answer 1

Probability that in a sample of 3 mobiles; 2 will be defective mobile is 0.096.

Step-by-step explanation:

We are given that a box contains 30 mobiles 6 of which are defective mobile.

A sample of 3 mobile are chosen at random from the box.

The above situation can be represented through binomial distribution;

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

where, n = number of trials (samples) taken = 3 mobiles

            r = number of success = 2 defective mobile

            p = probability of success which in our question is probability

                   that mobile is defective, i.e; p =  $\frac{6}{30}$  = 20%

Let X = Number of defective mobiles

SO, X ~ Binom(n = 3, p = 0.20)

Now, probability that in a sample of 3 mobiles 2 will be defective mobile is given by = P(X = 2)

               P(X = 2) =  $\binom{3}{2} \times 0.20^{2}\times (1-0.20)^{3-2}$

                             =  $3 \times 0.20^{2} \times 0.80^{1}$

                             =  0.096

Therefore, probability that in a sample of 3 mobiles; 2 will be defective mobile is 0.096.

Answer 2

Step-by-step explanation:

not correct