In a communication network, 98 % of messages | vedclass

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(A) Given that the total number of messages $n = 500$ and the probability of a message being transmitted correctly is $0.98$. Therefore, the probabili

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$\frac{11}{e^{10}}$

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(A) Given that the total number of messages $n = 500$ and the probability of a message being transmitted correctly is $0.98$. Therefore, the probability of a message being transmitted incorrectly is $p = 1 - 0.98 = 0.02$. Since $n$ is large and $p$ is small, we use the Poisson distribution with parameter $\lambda = np = 500 	imes 0.02 = 10$. The probability of $X$ incorrectly transmitted messages is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$. We need to find the probability that at most one message is transmitted incorrectly, i.e.,$P(X \leq 1) = P(X=0) + P(X=1)$. $P(X=0) = \frac{10^0 e^{-10}}{0!} = e^{-10}$. $P(X=1) = \frac{10^1 e^{-10}}{1!} = 10 e^{-10}$. Thus, $P(X \leq 1) = e^{-10} + 10 e^{-10} = 11 e^{-10} = \frac{11}{e^{10}}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

In a communication network, $98 \%$ of messages are transmitted with no error. If a random variable $X$ denotes the number of incorrectly transmitted messages, then the probability that at most one message is transmitted incorrectly out of $500$ messages sent, is

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