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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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$1-(0.95)^{5}$

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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 a bulb fusing is $p=0.05$.Therefore,the probability of 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 $x=0, 1, 2, 3, 4, 5$.We need to find the probability that at least one bulb fuses,which is $P(X \geq 1)$.$P(X \geq 1) = 1 - P(X Using the formula for $x=0$:$P(X=0) = ^{5}C_{0} (0.95)^{5-0} (0.05)^{0} = 1 	imes (0.95)^{5} 	imes 1 = (0.95)^{5}$.Therefore,$P(X \geq 1) = 1 - (0.95)^{5}$.

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,at least one will fuse after $150$ days.

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