SOLUTION: Experience has shown that 9/200 of all CDs produced by a certain machine are defective. If a quality control technician randomly tests twenty CDs, compute each of the following
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-> SOLUTION: Experience has shown that 9/200 of all CDs produced by a certain machine are defective. If a quality control technician randomly tests twenty CDs, compute each of the following
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Question 1105974: Experience has shown that 9/200 of all CDs produced by a certain machine are defective. If a quality control technician randomly tests twenty CDs, compute each of the following probabilities:
i- P(exactly one is defective)
ii- P(half are defective)
iii- P( no more than two are defective)
You can put this solution on YOUR website! Use the Binomial probability distribution
:
Probability (P) ( k successes in n trials) = nCk * p^k * (1-p)^(n-k), where nCk = n! / (k! * (n-k)!)
:
we know that p = 9/200 = 0.045, n = 20
:
i) P ( 1 is defective out of 20 ) = 20C1 * (0.045)^1 * (1-0.045)^(20-1) = 0.375
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ii) P ( 10 are defective out of 20 ) = 20C10 * (0.045)^10 * (1-0.045)^(20-10) = 0.000
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iii) P( no more than two are defective) = P ( 0 are defective out of 20 ) + P ( 1 is defective out of 20 ) + P ( 2 are defective out of 20 ) = 0.398 + 0.375 + 0.168 = 0.941
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