In a hospital,on an average if there are 35 b | vedclass

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(B) Given,the average number of births in a week is $35$. Since there are $7$ days in a week,the average number of births in a day is $\lambda = \frac

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

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(B) Given,the average number of births in a week is $35$. Since there are $7$ days in a week,the average number of births in a day is $\lambda = \frac{35}{7} = 5$. We use the Poisson distribution formula $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$. We need to find the probability of less than $3$ births in a day,which is $P(X $P(X = 0) = \frac{5^0 e^{-5}}{0!} = e^{-5}$. $P(X = 1) = \frac{5^1 e^{-5}}{1!} = 5e^{-5}$. $P(X = 2) = \frac{5^2 e^{-5}}{2!} = \frac{25}{2}e^{-5}$. Adding these probabilities: $P(X < 3) = e^{-5} + 5e^{-5} + \frac{25}{2}e^{-5} = e^{-5}(1 + 5 + 12.5) = 18.5e^{-5} = \frac{37}{2e^5}$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

In a hospital,on an average if there are $35$ births in a week,then the probability that there will be less than $3$ births in a day is:

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