Question

A manufacturer knows that 14% of the products it produces are defective. Find the probability that among 7 randomly selected products, at least one of them is defective.

Answer 1

Answer: 0.6521

Step-by-step explanation:

According to the Binomial distribution , the provability of getting x successes in n trials is given by :-

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where p=probability of getting success in each trial.

Let x denotes the number of defective products.

here , n=7 and p =0.14

Then, the probability that among 7 randomly selected products, at least one of them is defective= P(X ≥ 1) =1- P(X<1)

= 1- P(X=0)

$=1-^7C_0(0.14)^0(1-0.14)^7$

$=1-(1(1)(0.86)^7\ \[\because ^nC_0=1]$

$=1-0.347927822217$

$=0.652072177783\approx0.6521$

Hence, the probability that among 7 randomly selected products, at least one of them is defective is 0.6521 .