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A fair coin is tossed three times. Let $X$ be the number of heads that turn up on the first two tosses and $Y$ the number of heads that turn up on the third toss. Give the distribution of $X$, $Y$, $X + Y$, $X − Y$ and $XY$.

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    $\begingroup$ What are your thoughts? What have you tried? $\endgroup$ – Sloan Apr 19 '15 at 15:24
  • $\begingroup$ $X=\{0,1,2\} Y=\{0,1\}$ $\endgroup$ – Karolina Sz Apr 19 '15 at 15:48
  • $\begingroup$ $P(Y=1)=1/2, P(Y=0)=1/2, P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4$ $\endgroup$ – Karolina Sz Apr 19 '15 at 15:59
  • $\begingroup$ you didn't told what you did or your problem with X+Y,X-Y $\endgroup$ – RE60K Apr 19 '15 at 16:33
  • $\begingroup$ So is my answer ok ? $\endgroup$ – Karolina Sz Apr 19 '15 at 16:42
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Is it good answer? $$X=\{0,1,2\}, Y=\{0,1\}$$ $$P(Y=1)=1/2, P(Y=0)=1/2, P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4$$ $$P(X+Y=0)=1/8, P(X+Y=1)=3/8, P(X+Y=2)=3/8, P(X+Y=3)=1/8$$ $$P(X-Y=-1)=1/8, P(X-Y=0)=3/8, P(X-Y=1)=3/8, P(X-Y=2)=1/8$$ $P(XY=0)=5/8, P(XY=1)=2/8, P(XY=2)=1/8$

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  • $\begingroup$ The probabilities are ok. You can write them in a table, like $$\begin{array}{|c|c|c|} \hline X+Y & 0& 1 & 2 & 3 \\ \hline P(X+Y) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array}$$ $\endgroup$ – callculus Apr 19 '15 at 17:15
  • $\begingroup$ But isn´t $P(X*Y=1)=\frac{2}{8}$ ? There are two combination (1,0,1) and (0,1,1). And for $P(X*Y=0)$ I have $\frac{5}{8}$ $\endgroup$ – callculus Apr 19 '15 at 17:24
  • $\begingroup$ you right, I corrrect it $\endgroup$ – Karolina Sz Apr 19 '15 at 17:26
  • $\begingroup$ Then it will be fine. $\endgroup$ – callculus Apr 19 '15 at 17:26
  • $\begingroup$ It looked like the OP asked for the distribution of $X-Y$, not $XY$. $\endgroup$ – Brian Tung Apr 19 '15 at 17:46
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The distribution of $X+Y$ is a binomial distribution, as expected:

$$ q_0 = q_3 = 1/8 \\ q_1 = q_2 = 3/8 $$

where $q_i = P(X+Y = i)$. This can be reasoned out as follows: You can obtain $0$ only if $X = Y = 0$, so the probability is $(1/4)(1/2) = 1/8$. You can obtain $3$ only if $X = 2, Y = 1$, so again the probability is $(1/4)(1/2) = 1/8$. Finally, you can obtain $1$ if $X = 1, Y = 0$, or vice versa, and you can obtain $2$ if $X = 2, Y = 0$ or $X = Y = 1$, and in either case the probability is $(1/4)(1/2)+(1/2)(1/2) = 3/8$.

Interestingly, the difference $X-Y$ has the same spread, only shifted down by one:

$$ r_{-1} = r_2 = 1/8 \\ r_0 = r_1 = 3/8 $$

where $r_i = P(X-Y = i)$. Can you see, by a similar consideration of the different possibilities, why that is?

ETA: Ahh, OK, for $XY$, again, a similar consideration of cases gives us

$$ p_0 = P(X = \mbox{anything}, Y = 0) + P(X = 0, Y = 1) = 1/2 + (1/4)(1/2) = 5/8 \\ p_1 = P(X = Y = 1) = (1/2)(1/2) = 1/4 \\ P_2 = P(X = 2, Y = 1) = (1/4)(1/2) = 1/8 $$

where $p_i = P(XY = i)$.

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