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(B) Let $X$ be the number of calls made per hour. Since the average number of calls is $5$, $X$ follows a Poisson distribution with parameter $\lambda

Vedclass · Accepted Answer

$\frac{2 e^5-37}{2 e^5}$

Vedclass · Answer

(B) Let $X$ be the number of calls made per hour. Since the average number of calls is $5$, $X$ follows a Poisson distribution with parameter $\lambda = 5$.The cost of $X$ calls is $2X$. We want to find the probability that the cost exceeds $Rs. 4$, i.e.,$P(2X > 4) = P(X > 2)$.$P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.Using the Poisson formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$:$P(X=0) = e^{-5} \frac{5^0}{0!} = e^{-5}$.$P(X=1) = e^{-5} \frac{5^1}{1!} = 5e^{-5}$.$P(X=2) = e^{-5} \frac{5^2}{2!} = \frac{25}{2} e^{-5}$.Summing these: $P(X \leq 2) = e^{-5} (1 + 5 + 12.5) = 18.5 e^{-5} = \frac{37}{2} e^{-5}$.Therefore, $P(X > 2) = 1 - \frac{37}{2 e^5} = \frac{2 e^5 - 37}{2 e^5}$.

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

An executive in a company makes on an average $5$ telephone calls per hour at a cost of $Rs. 2$ per call. The probability that in any hour the cost of the calls exceeds a sum of $Rs. 4$ is

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