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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.

User Sanan Ali
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Answer: 0.6521

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 .

User Jeffalstott
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