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(B) Let $X$ be the number of bulbs that fuse after $150$ days. This follows a binomial distribution with $n = 5$ and $p = 0.05$.Here,$q = 1 - p = 0.95

Vedclass · Accepted Answer

$1 - (0.95)^4 	imes 1.25$

Vedclass · Answer

(B) Let $X$ be the number of bulbs that fuse after $150$ days. This follows a binomial distribution with $n = 5$ and $p = 0.05$.Here,$q = 1 - p = 0.95$.The probability mass function is $P(X = x) = ^nC_x q^{n-x} p^x = ^5C_x (0.95)^{5-x} (0.05)^x$.We need to find the probability that more than one bulb fuses,i.e.,$P(X > 1)$.$P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)]$.$P(X = 0) = ^5C_0 (0.95)^5 (0.05)^0 = (0.95)^5$.$P(X = 1) = ^5C_1 (0.95)^4 (0.05)^1 = 5 	imes (0.95)^4 	imes 0.05 = 0.25 	imes (0.95)^4$.Therefore,$P(X > 1) = 1 - [(0.95)^5 + 0.25(0.95)^4]$.Simplifying,$P(X > 1) = 1 - (0.95)^4 [0.95 + 0.25] = 1 - 1.2(0.95)^4$.

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,more than one will fuse after $150$ days.

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