A fair coin is tossed four times. Let the random variable $X$ denote the number of heads in the first 3 tosses, and let the random variable $Y$ denote the number of heads in the last 3 tosses. What is the joint pmf of $X$ and $Y$?
The solution is the following table, which I am having trouble with.
$$
\begin{array}{l|llll}
X \backslash Y
& 0
& 1
& 2
& 3
\\
\hline
0
& 1/16
& 1/16
& 0
& 0
\\
1
& 1/16
& 3/16
& 2/16
& 0
\\
2
& 0
& 2/16
& 3/16
& 1/16
\\
3
& 0
& 0
& 1/16
& 1/16
\end{array}
$$
(Original image here).
I do not understand $(1,1)$. Where do they get $3/16$ from? The only possibility of only $1$ head in the first $3$ tosses and only $1$ in the last $3$ tosses is HTTH, hence it should be $1/16$?
Furthermore I do not understand $(2,2)$. Where do they get $3/16$ from? The only possibility of only $2$ heads in both the first $3$ tosses and the last $3$ tosses is THHT, hence it should also be $1/16$?
Best Answer
For $(1,1)$, note that in contrary to what you are writing, $$ \{X=1, Y=1\} = \{HTTH, THTT, TTHT\} $$ and for $(2,2)$ the same trick works $$ \{X=2, Y=2\} = \{THHT, HHTH, HTHH\} $$