Question

The probability that a machine part is defective is 0.1. Find the probability that no more than 2 out of 12 parts tested are defective.

Answer 1

Answer:

Probability that no more than 2 out of 12 parts tested are defective is 0.8891.

Step-by-step explanation:

We are given that the probability that a machine part is defective is 0.1.

Twelve parts are selected at random.

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 trials (samples) taken = 12 parts

            r = number of success = no more than 2

            p = probability of success which in our question is probability that

                  a machine part is defective, i.e; p = 0.1

Let X = Number of machine parts that are defective

So, X ~ Binom(n = 12, p = 0.1)

Now, Probability that no more than 2 out of 12 parts tested are defective is given by = P(X $\leq$ 2)

P(X $\leq$ 2) =  P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{12}{0} \times 0.1^{0} \times (1-0.1)^{12-0}+\binom{12}{1} \times 0.1^{1} \times (1-0.1)^{12-1}+\binom{12}{2} \times 0.1^{2} \times (1-0.1)^{12-2}$

=   $1\times 1 \times 0.9^{12}+ 12 \times 0.1^{1} \times 0.9^{11}+ 66\times 0.1^{2} \times 0.9^{10}$

=  0.8891

Therefore, probability that no more than 2 out of 12 parts tested are defective is 0.8891.