The items produced by a firm are supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defective items?

Solution

Here n = 8
$straight p space equals space 5 over 100 space equals 1 over 20 straight q space equals space 1 minus space straight p space equals space 1 space minus space 1 over 20 space equals 19 over 20$
P(less than 2 defective )
= P(0) + P(1)
$equals space straight C presuperscript 8 subscript 0 space open parentheses 19 over 20 close parentheses to the power of 8 minus 0 end exponent space open parentheses 1 over 20 close parentheses to the power of 0 space space plus straight C presuperscript 8 subscript 1 space open parentheses 19 over 20 close parentheses to the power of 8 minus 1 end exponent space open parentheses 1 over 20 close parentheses to the power of 1 equals space 1 space cross times space open parentheses 19 over 20 close parentheses to the power of 8 space cross times space 1 space plus space 8 over 1 cross times open parentheses 19 over 20 close parentheses to the power of 7 space cross times 1 over 20 equals space open parentheses 19 over 20 close parentheses to the power of 7 space open square brackets 19 over 20 plus 8 over 20 close square brackets space equals space open parentheses 19 over 20 close parentheses to the power of 9 space cross times space 27 over 20$

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Probability

The items produced by a firm are supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defective items?

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