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\begin{questionparts}
\item $\,$
\begin{align*}
&& \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) &= \binom{30}{1} \frac{1}{36} \left ( \frac{35}{36} \right)^{29} \binom{30}{0} \left ( \frac{24}{25} \right)^{30} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\
&&&= \frac{30 \cdot 35^{29} \cdot 24^{30} \cdot 40^{30}}{36^{30} \cdot 25^{30} \cdot {41}^{30}}\\
&&&= \frac{1}{35} N\\
&& \mathbb{P}(\text{B took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{1} \frac{1}{25} \left ( \frac{24}{25} \right)^{29} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\
&&&= \frac{1}{24} N \\
&& \mathbb{P}(\text{C took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{0}\left ( \frac{24}{25} \right)^{30} \binom{30}{1} \frac{1}{41} \left ( \frac{40}{41} \right)^{29}\\
&&&= \frac{1}{40} N \\
&& \mathbb{P}(\text{Arthur took wicket} | \text{exactly one wicket}) &= \frac{ \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) }{ \mathbb{P}(\text{exactly one wicket}) } \\
&&&= \frac{ \frac{1}{35} N}{\frac1{35} N + \frac{1}{24}N + \frac{1}{40} N} \\
&&&= \frac{3}{10}
\end{align*}

Alternatively, we could look at:

\begin{align*}
&& \mathbb{P}(X_A = 1 | X_A + X_B + X_C =1) &= \frac{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)}{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 1,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 0,X_C = 1)} \\
&&&= \frac{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}}{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}+\frac{\mathbb{P}(X_B = 1)}{\mathbb{P}(X_B=0)}+\frac{\mathbb{P}(X_C = 1)}{\mathbb{P}(X_C=0)}}
\end{align*}

and we can calculate these relatively likelihoods in a similar way to above.

\item $\,$ \begin{align*}
&& \mathbb{E}(\text{number of wickets}) &= \mathbb{E} \left ( \sum_{i=1}^{90} \mathbb{1}_{i\text{th ball is a wicket}} \right) \\
&&&=  \sum_{i=1}^{90} \mathbb{E} \left (\mathbb{1}_{i\text{th ball is a wicket}} \right) \\
&&&= 30 \cdot \frac{1}{36} + 30 \cdot \frac{1}{25}  + 30 \cdot \frac{1}{41} \\
&&&\approx 1 + 1 + 1 = 3
\end{align*}

We might model the number of wickets taken as $Po(\lambda)$, where $\lambda$ is the average number of wickets taken. We can think of this roughly as the Poisson approximation to the binomial where $N$ is large and $Np$ is small.

Assuming we use $Po(3)$ we have

\begin{align*}
&& \mathbb{P}(\text{at least 5 wickets}) &= 1-\mathbb{P}(\text{4 or fewer wickets}) \\
&&&= 1- e^{-3} \left (1 + \frac{3}{1} + \frac{3^2}{2} + \frac{3^3}{6} + \frac{3^4}{24} \right) \\
&&&= 1 - \frac{1}{20} \left ( 1 + 3 + \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) \\
&&&= 1 - \frac{1}{20} \left (13 + 3\tfrac38 \right) \\
&&&\approx 1 - \frac{16}{20} = \frac15
\end{align*}

\end{questionparts}

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