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Question
In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?
Solution
Let X denote the number of defective items in a sample of 10 items.
X follows a binomial distribution with n =10;
p = probability of defective items = 5 % = 0 . 05 ; q = 1 - p = 0 . 95
\[P(X = r) = ^{10}{}{C}_r (0 . 05 )^r (0 . 95 )^{10 - r} \]
\[\text{ Probability ( sample of 10 items will include not more than one defective item)} = P(X \leq 1) \]
\[ = P(X = 0) + P(X = 1) \]
\[ = ^{10}{}{C}_0 (0 . 05 )^0 (0 . 95 )^{10 - 0} + ^ {10}{}{C}_1 (0 . 05 )^1 (0 . 95 )^{10 - 1} \]
\[ = ( 0 . 95 )^9 (0 . 95 + 0 . 5)\]
\[ = 1 . 45(0 . 95 )^9\]
\[= \left( \frac{19}{20} \right)^9 \frac{29}{20}; ( \because 1 . 45 = \frac{29}{20}\text{ and } 0 . 95 = \frac{19}{20})\]
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