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Question
There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.
Solution
\[\text{ Let X denote the number of defective items ina sample of 8 items . Then, X follows a binomial distribution with n = 8, } \]
\[p = (\text{ probability of getting a defective item} ) = 0 . 06 \text{ and } q = 1 - p = 0 . 94\]
\[P(X = r) = ^{8}{}{C}_r (0 . 06 )^r (0 . 94 )^{8 - r} , r = 0, 1, 2, 3, . . . 8\]
\[\text{ The required probability = Probability of not more than one defective item } \]
\[ = P (X \leq 1) \]
\[ = P(X = 0) + P(X = 1)\]
\[=^{8}{}{C}_0 (0 . 06 )^0 (0 . 94 )^{8 - 0} + ^{8}{}{C}_1 (0 . 06 )^1 (0 . 94 )^{8 - 1} \]
\[ = (0 . 94 )^8 + 8(0 . 06)(0 . 94 )^7 \]
\[ = (0 . 94 )^7 \left\{ 0 . 94 + 0 . 48 \right\}\]
\[ = 1 . 42(0 . 94 )^7\]
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