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$$X\sim\mathrm{binomial}(1, 1/3)\text{ and }Y\sim\mathrm{binomial}(2,1/2)$$

How can I get

$$W = XY+1$$

Normally I would attempt but this one I don't even know how to get started

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  • $\begingroup$ $X$ only has two outcomes, $Y$ has three. Draw a table with $X$'s outcomes going down the side and $Y$'s across the top. Then at their intersections write their product and add $1$. This gives you the sample space for $W$. Can you figure out the product of each outcome? I assume $X$ and $Y$ are independent here. $\endgroup$ – Patrick Oct 25 '13 at 0:30
  • $\begingroup$ How did you know that X only has two outcomes and Y has three? $\endgroup$ – user2472706 Oct 25 '13 at 0:33
  • $\begingroup$ Usually, when a random variable $N\sim\mathbb{Bin}(n,p)$ then we say $N$ counts the number of successes after performing an independent experiment $n$ times. This means that there could be $0, 1, 2, ..., n$ possible successes or $n+1$ outcomes. $\endgroup$ – Patrick Oct 25 '13 at 0:39
  • $\begingroup$ So the answer is basically a 2 by 3 table, with each tiles giving the probability given some X, Y, am i correct? And that probabiity coming from the product of [p(x)p(y)]+1 $\endgroup$ – user2472706 Oct 25 '13 at 0:47
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$$ \begin{array}{c|ccc|c} X \backslash Y & \ 0 & \ 1 & \ 2 & \ \\ \hline 0 & 1 & 1 & 1 & \ 2/3 \\ 1 & 1 & \fbox{2} & 3 & \fbox{1/3} \\ \hline & \ (1/2)^2 & \ \fbox{$2(1/2)^2$} & \ (1/2)^2 & \mathbb{Pr}(Y) \backslash \mathbb{Pr}(X) \\ \end{array} $$

From the above table we see that $\mathbb{Pr}(W=2) = \mathbb{Pr}(X=1,Y=1) = 2(1/2)^2(1/3)=1/6.$

Can you do the rest?

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