Show that, as the number of trials $n$ becomes large but $n p_{i}=\lambda_{i}, i=1,2, \ldots, k-1$ remains finite, the multinomial probability distribution (26.146),
$$
M_{n}\left(x_{1}, x_{2}, \ldots, x_{k}\right)=\frac{n !}{x_{1} ! x_{2} ! \cdots x_{k} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}}
$$
can be approximated by a multiple Poisson distribution (with $k-1$ factors)
$$
M_{n}^{\prime}\left(x_{1}, x_{2}, \ldots, x_{k-1}\right)=\prod_{i=1}^{k-1} \frac{e^{-\lambda_{i}} \lambda_{i}^{x_{i}}}{x_{i} !}
$$
(Write $\sum_{i}^{k-1} p_{i}=\delta$ and express all terms involving subscript $k$ in terms of $n$ and $\delta$, either exactly or approximately. You will need to use $n ! \approx n^{f}[(n-\epsilon) !]$ and $(1-a / n)^{n} \approx e^{-a}$ for large $\left.n_{1}\right)$
(a) Verify that the terms of $M_{n}^{\prime}$ when summed over all values of $x_{1}, x_{2}, \ldots, x_{k-1}$ add up to unity.
(b) If $k=7$ and $\lambda_{i}=9$ for all $i=1,2, \ldots, 6$, estimate, using the appropriate Gaussian approximation, the chance that at least three of $x_{1}, x_{2}, \ldots, x_{6}$ will be 15 or greater.