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

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$(0.95)^{4} 	imes 1.2$

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(A) Let $X$ represent the number of bulbs that will fuse after $150$ days of use in an experiment of $5$ trials. The trials are Bernoulli trials.It is given that $p = 0.05$.$	herefore q = 1 - p = 1 - 0.05 = 0.95$.$X$ follows a binomial distribution with $n = 5$ and $p = 0.05$.$	herefore P(X = x) = ^{n}C_{x} q^{n-x} p^{x}$,where $x = 0, 1, 2, ..., 5$.$= ^{5}C_{x} (0.95)^{5-x} \cdot (0.05)^{x}$.$P(	ext{not more than one}) = P(X \leq 1) = P(X = 0) + P(X = 1)$.$= ^{5}C_{0} 	imes (0.95)^{5} \cdot (0.05)^{0} + ^{5}C_{1} (0.95)^{4} \cdot (0.05)^{1}$.$= 1 	imes (0.95)^{5} + 5 	imes (0.95)^{4} 	imes 0.05$.$= (0.95)^{5} + 0.25 	imes (0.95)^{4}$.$= (0.95)^{4} 	imes [0.95 + 0.25]$.$= (0.95)^{4} 	imes 1.2$.

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,not more than one will fuse?

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