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Question
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs not more than one will fuse after 150 days of use
Solution
Let X be the number of bulbs that fuse after 150 days.
X follows a binomial distribution with n = 5,
\[\text{ Or } p = \frac{1}{20}\text{ and } q = \frac{19}{20}\]
\[P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r} \]
\[\text{ Probability (not more than 1 will fuse after 150 days of use } ) = P(X \leq 1) \]
\[ = P(X = 0) + P(X = 1) \]
\[ = \left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1} \]
\[ = \left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\} \]
\[ = \frac{6}{5} \left( \frac{19}{20} \right)^4 \]
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