Question

Before a television set leaves the factory, it is given a quality control check. The probability that a television contains 0, 1, or 2 defects is 0.88, 0.08, and 0.04, respectively. In a sample of 16 televisions, find the probability that 9 will have 0 defects, 4 will have 1 defect, and 3 will have 2 defects

Answer 1

Answer:

Probability that 9 will have 0 defects is 0.0013

Probability that 4 will have 1 defect is 0.0274

Probability that 3 will have 2 defects is 0.0211

Step-by-step explanation:

We are given that before a television set leaves the factory, it is given a quality control check. The probability that a television contains 0, 1, or 2 defects is 0.88, 0.08, and 0.04, respectively.

A sample of 16 televisions 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 = 16 televisions

             r = number of success

            p = probability of success which in our question is probability

                 of television containing defects

LET X = Number of televisions containing defects

(a) Now, in first part we have to find the probability that 9 will have 0 defects.

Here, p = probability that television contain 0 defects = 0.88

and r = 9

SO, X ~ Binom(n = 16, p = 0.88)

Probability that 9 will have 0 defects is given by = P(X = 9)

                   P(X = 9) = $\binom{16}{9} \times 0.88^{9} \times (1-0.88)^{16-9}$

                                 = $11440 \times 0.88^{9}\times 0.12^{7}$

                                 = 0.0013

(b) Now, in second part we have to find the probability that 4 will have 1 defect.

Here, p = probability that television contain 1 defect = 0.08

and r = 4

SO, X ~ Binom(n = 16, p = 0.08)

Probability that 4 will have 1 defect is given by = P(X = 4)

                   P(X = 4) = $\binom{16}{4} \times 0.08^{4} \times (1-0.08)^{16-4}$

                                 = $1820 \times 0.08^{4}\times 0.92^{12}$

                                 = 0.0274

(c) Now, in third part we have to find the probability that 3 will have 2 defects.

Here, p = probability that television contain 2 defects = 0.04

and r = 3

SO, X ~ Binom(n = 16, p = 0.04)

Probability that 3 will have 2 defects is given by = P(X = 3)

                   P(X = 3) = $\binom{16}{3} \times 0.04^{3} \times (1-0.96)^{16-3}$

                                 = $560 \times 0.04^{3}\times 0.96^{13}$

                                 = 0.0211