A contestant participates in a game show where three important prizes are offered. A contestant participates in a game show where three important prizes are offered. His chances of winning the three prizes are $\frac{1}{6}$, $\frac{1}{3}$ and $\frac{1}{2}$, respectively. 


*

*How many prizes can the contestant expect to win?

*Let $X$ denote the number of prizes won by the contestant. Find the
probability distribution function of $X$.


For 1. Using the Poisson model I obtained that $$P(0)=\frac{10}{36}, P(1)=\frac{17}{36}, P(2)=\frac{8}{36}, P(3)=\frac{1}{36}$$ and I think that the correct answer is 1 because it has the highest probability. 
Is it correct?
For 2. Is it alright if I write directly $$
\begin{pmatrix}
0 & 1 & 2 & 3 \\
\frac{10}{36} & \frac{17}{36} & \frac{8}{36} & \frac{1}{36}\\
\end{pmatrix}
$$

$X$ does not have a Poisson distribution because it only takes finitely many values. Observe that $X=X_1+X_2+X_3$ where $X_1\sim\mathrm{Ber}\left(\frac16\right)$, $X_2\sim\mathrm{Ber}\left(\frac13\right)$, and $X_3\sim\mathrm{Ber}\left(\frac12\right)$ are independent. Therefore
\begin{align}
\mathbb P(X=0) &= \mathbb P(X_1=0,X_2=0,X_3=0)\\ &= \mathbb P(X_1=0)\mathbb P(X_2=0)\mathbb P(X_3=0)\\
&=\left(1-\frac16\right)\left(1-\frac13\right)\left(1-\frac12\right)\\
&= \frac5{18},
\end{align}
\begin{align}
\mathbb P(X=1) &= \mathbb P\left(\bigcup_{i=1}^3 \{X_i=1\}\cap\bigcap_{j\in\{1,2,3\}\setminus\{i\}}\{X_j=0\} \right)\\
&= \sum_{i=1}^3 \mathbb P(X_i=1)\prod_{j\in\{1,2,3\}\setminus i\}}\mathbb P(X_j=0)\\
&= \mathbb P(X_1=1)\mathbb P(X_2=0)\mathbb P(X_3=0)+\mathbb P(X_1=0)\mathbb P(X_2=1)\mathbb P(X_3=0)+\mathbb P(X_1=0)\mathbb P(X_2=0)\mathbb P(X_3=1)\\
&= \frac16\left(1-\frac13\right)\left(1-\frac12\right) +\left(1-\frac16\right)\frac13\left(1-\frac12\right) + \left(1-\frac16\right)\left(1-\frac13\right)\frac12 \\&= \frac{17}{36},
\end{align}
\begin{align}
\mathbb P(X=2) &= \mathbb P(X_1+X_2+X_3=2)\\
&= \mathbb P(X_1=1)\mathbb P(X_2=1)\mathbb P(X_3=0)+\mathbb P(X_1=1)\mathbb P(X_2=0)\mathbb P(X_3=1)+\mathbb P(X_1=0)\mathbb P(X_2=1)\mathbb P(X_3=1)\\
&= \frac16\cdot\frac13\left(1-\frac12\right)+ \frac16\left(1-\frac13\right)\frac12 + \left(1-\frac16\right)\frac13\cdot\frac12\\
&= \frac29,
\end{align}
and
\begin{align}
\mathbb P(X=3) &= \mathbb P(X_1+X_2+X_3=3)\\ &= \mathbb P(X_1=1)\mathbb P(X_2=1)\mathbb P(X_3=1)\\
&= \frac16\cdot\frac13\cdot\frac12 = \frac1{36}.
\end{align}
The expected value of $X$ is
\begin{align}
\mathbb E[X] &= \sum_{i=0}^3 i\cdot\mathbb P(X=i)\\
&= 0\cdot\frac5{18} + 1\cdot\frac{17}{36} + 2\cdot\frac29 + 3\cdot\frac1{36}\\
&= 1.
\end{align}
