Question

A packet of 10 electronic components is known to include 2 defectives. If a sample of 4 components is selected at random from the packet, what is the probability that the sample does not contain more than 1 defective?​

Answer 1

The probability that the sample does not contain more than 1 defective is 0.8192.

Step-by-step explanation:

We are given that a packet of 10 electronic components is known to include 2 defectives.

Also, a sample of 4 components is selected at random from the packet.

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 = 4 electric components

            r = number of success = not more than 1 defective

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

                electric components are defective, i.e; (2÷10) = 0.20 or 20%

LET X = Number of electric components that are defective

SO, X ~ Binom(n = 4, p = 0.20)

Now, probability that the sample does not contain more than 1 defective is given by = P(X $\leq$ 1)

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

                =  $\binom{4}{0}\times 0.20^{0}\times (1-0.20)^{4-0} +\binom{4}{1}\times 0.20^{1}\times (1-0.20)^{4-1}$

                =  $1 \times 1 \times 0.80^{4} + 4 \times 0.20 \times 0.80^{3}$

                =  0.4096 + 0.4096

                =  0.8192

Therefore, probability that the sample does not contain more than 1 defective is 0.8192.