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## Fill in the blanks. GATE-Biotechnology-2013

One percent of the cars manufactured by a company are defective. What is the probability (upto four decimals) that more than two cars are defective, if 100 cars are produced? __________

By:Fututron

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## Answers

This problem uses the binomial probability formula,

Pr( k success in n trials) = nCk * p^k * q^(n-k)

We are given a sample size of 100 cars and told that 1% of the cars are defective and asked what is the probability of more than 2 cars being defective

let's look at a success as failure of a car, then p = .01 and q = 0.99

Pr( k > 2 ) = 1 - Pr(k=0) - Pr(k=1) - Pr(k=2)

Pr(k=0) = 100C0 * (.01)^0 * (.99)^(100-0) = 0.3660323412732289

Pr(k=1) = 100C1 * (.01)^1 * (.99)^(100-1) = 0.36972963764972616

Pr(k=2) = 100C2 * (.01)^2 * (.99)^(100-2) = 0.18486481882486308

Pr(k > 2 ) = 1 − 0.920626798 = 0.079373202 approx 0.0794

Amogh

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