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(B) The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{60}{600} = 0.1$. The probability mass function is giv

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$\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)$

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(B) The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{60}{600} = 0.1$. The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-0.1} (0.1)^x}{x!}$. We need to find the probability that a page contains at most two errors, which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. $P(X=0) = \frac{e^{-0.1} (0.1)^0}{0!} = e^{-0.1}$. $P(X=1) = \frac{e^{-0.1} (0.1)^1}{1!} = 0.1 e^{-0.1}$. $P(X=2) = \frac{e^{-0.1} (0.1)^2}{2!} = \frac{0.01}{2} e^{-0.1} = 0.005 e^{-0.1}$. Adding these probabilities: $P(X \le 2) = e^{-0.1} (1 + 0.1 + 0.005) = e^{-0.1} (1.105) = e^{-0.1} \left(\frac{1105}{1000}\right) = e^{-0.1} \left(\frac{221}{200}\right) = \frac{1}{e^{0.1}} \left(\frac{221}{200}\right)$.

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

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

In a book consisting of $600$ pages, there are $60$ typographical errors. The probability that a randomly chosen page will contain at most two errors is

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