SOLUTION: Q : One percent of the cars manufactured by a company are defective. What is the probability ( up to four decimals ) that more than two cars are defective, if 100 cars are produced
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-> SOLUTION: Q : One percent of the cars manufactured by a company are defective. What is the probability ( up to four decimals ) that more than two cars are defective, if 100 cars are produced
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Question 1015429: Q : One percent of the cars manufactured by a company are defective. What is the probability ( up to four decimals ) that more than two cars are defective, if 100 cars are produced. ??
help me to solve this problem also want some hint, the way to solve defective kind a problems. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! 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
