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 

Answer 1

Let X be the number of bulbs that fuse after 150 days.
X follows a binomial distribution with n = 5,

\[p = 0 . 05 \text{ and }  q = 0 . 95\]

\[\text{ Or } p = \frac{1}{20}\text{ and } q = \frac{19}{20}\]

\[P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r} \]

\[\text{ Probability (not more than 1 will fuse after 150 days of use } ) = P(X \leq 1) \]

\[ = P(X = 0) + P(X = 1) \]

\[ = \left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1} \]

\[ = \left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\} \]

\[ = \frac{6}{5} \left( \frac{19}{20} \right)^4 \]