Question

A manufacturing process turns out articles that are on the average 10% defective. Compute the probability of 0,1,2 and 3 defective articles that might occur in a sample of 3 articles.

Answer 1

P(X = 0) = 0.729

P(X = 1) = 0.243

PX = 2) = 0.027

P(X = 3) = 0.001

Step-by-step explanation:

We are given that a manufacturing process turns out articles that are on the average 10% defective.

Also, a sample of 3 articles is selected.

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 = 3 articles

            r = number of success

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

               articles are defective, i.e; 10%

LET X = Number of articles that are defective

SO, X ~ Binom(n = 3, p = 0.10)

(a) Probability that there are 0 defective articles in a sample of 3 articles is given by = P(X = 0)

                     P(X = 0) =  $\binom{3}{0}\times 0.10^{0}\times (1-0.10)^{3-0}$

                                   =  $1\times 1 \times (0.90)^{3}$

                                   = 0.729

(b) Probability that there are 1 defective articles in a sample of 3 articles is given by = P(X = 1)

                     P(X = 1) =  $\binom{3}{1}\times 0.10^{1}\times (1-0.10)^{3-1}$

                                   =  $3\times 0.10 \times (0.90)^{2}$

                                   = 0.243

(c) Probability that there are 2 defective articles in a sample of 3 articles is given by = P(X = 2)

                     P(X = 2) =  $\binom{3}{2}\times 0.10^{2}\times (1-0.10)^{3-2}$

                                   =  $3\times 0.10^{2} \times (0.90)^{1}$

                                   = 0.027

(d) Probability that there are 3 defective articles in a sample of 3 articles is given by = P(X = 3)

                     P(X = 3) =  $\binom{3}{3}\times 0.10^{3}\times (1-0.10)^{3-3}$

                                   =  $1\times 0.10^{3} \times (0.90)^{0}$

                                   = 0.001