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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 none will fuse after 150 days of use
Sum
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Solution
Let X be the number of bulbs that fuse after 150 days.
X follows a binomial distribution with n = 5,
\[p = 0 . 05 \text{ and } q = 0 . 95\]
\[\text{ Or } p = \frac{1}{20}\text{ and } q = \frac{19}{20}\]
\[P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r} \]
\[\text{ Probability (none will fuse after 150 days of use } ) = P(X = 0) \]
\[ =^ {5}{}{C}_0 \left( \frac{1}{20} \right)^0 \left( \frac{19}{20} \right)^{5 - 0} \]
\[ = \left( \frac{19}{20} \right)^5 \]
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