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(B) Let $p$ be the probability that a bulb fuses,so $p = 0.05 = \frac{5}{100} = \frac{1}{20}$.Let $q$ be the probability that a bulb does not fuse,so

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${\left( {\frac{{19}}{{20}}} \right)^5}$

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(B) Let $p$ be the probability that a bulb fuses,so $p = 0.05 = \frac{5}{100} = \frac{1}{20}$.Let $q$ be the probability that a bulb does not fuse,so $q = 1 - p = 1 - \frac{1}{20} = \frac{19}{20}$.We are given $n = 5$ bulbs.We want the probability that none of the $5$ bulbs will fuse,which means all $5$ bulbs do not fuse.Using the binomial distribution formula $P(X = r) = {}^nC_r \cdot q^{n-r} \cdot p^r$,where $r = 0$ (number of fused bulbs).$P(X = 0) = {}^5C_0 \cdot (\frac{19}{20})^5 \cdot (\frac{1}{20})^0 = 1 \cdot (\frac{19}{20})^5 \cdot 1 = (\frac{19}{20})^5$.Thus,the correct option is $B$.

The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. What is the probability that out of $5$ such bulbs,none will fuse after $150$ days of use?

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