Find the distribution - coin is tossed three times 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 $XY$.
 A: Is it good answer? 
$$X=\{0,1,2\},  Y=\{0,1\}$$
$$P(Y=1)=1/2, P(Y=0)=1/2,  P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4$$
$$P(X+Y=0)=1/8, P(X+Y=1)=3/8, P(X+Y=2)=3/8, P(X+Y=3)=1/8$$
$$P(X-Y=-1)=1/8, P(X-Y=0)=3/8, P(X-Y=1)=3/8, P(X-Y=2)=1/8$$
$P(XY=0)=5/8, P(XY=1)=2/8, P(XY=2)=1/8$  
A: 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 $p_i = P(XY = i)$.
