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(A) Let $X$ represent the number of bulbs that will fuse after $150$ days of use in an experiment of $n=5$ trials. The trials are Bernoulli trials.It

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$(0.95)^{5}$

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(A) Let $X$ represent the number of bulbs that will fuse after $150$ days of use in an experiment of $n=5$ trials. The trials are Bernoulli trials.It is given that the probability of success (a bulb fusing) is $p=0.05$.Therefore,the probability of failure (a bulb not fusing) is $q=1-p=1-0.05=0.95$.$X$ follows a binomial distribution with $n=5$ and $p=0.05$.The probability mass function is given by $P(X=x) = ^{n}C_{x} q^{n-x} p^{x}$,where $x=0, 1, 2, ..., 5$.We need to find the probability that none of the bulbs fuse,which corresponds to $P(X=0)$.$P(X=0) = ^{5}C_{0} (0.95)^{5-0} (0.05)^{0}$.Since $^{5}C_{0} = 1$ and $(0.05)^{0} = 1$,we get:$P(X=0) = 1 	imes (0.95)^{5} 	imes 1 = (0.95)^{5}$.

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.

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