$\frac{1}{6};\frac{5}{10}$
Compute (a) $P(X=1|Y=1)$ and (b) $P(X = 2|Y=2)$ for the following joint distribution:
| $Y$ | $X=1$ | 2 | 3 |
|---|---|---|---|
| $1$ | $0.1$ | $0.2$ | $0.3$ |
| $2$ | $0.15$ | $0.15$ | $0$ |
| $3$ | $0.05$ | $0$ | $0.05$ |
$\frac{1}{6};\frac{5}{10}$
We first compute the marginal PMF $f_Y(y)$ of $Y$ from the given joint PMF $f_{XY}(x,y)$. The marginal PMF is given by the row sums of the given table.
| $Y$ | $X=1$ | 2 | 3 | $f_Y(y)$ |
|---|---|---|---|---|
| $1$ | $0.1$ | $0.2$ | $0.3$ | $0.6$ |
| $2$ | $0.15$ | $0.15$ | $0$ | $0.3$ |
| $3$ | $0.05$ | $0$ | $0.05$ | $0.1$ |
(a) Now, using the the values from the table we get,\[P(X=1|Y=1)=\frac{f_{XY}(1,1)}{f_Y(1)}=\frac{0.1}{0.6}=\frac{1}{6}.\]
(b) Now, using the the values from the table we get,\[P(X=2|Y=2)=\frac{f_{XY}(2,2)}{f_Y(2)}=\frac{0.15}{0.3}=\frac{5}{10}.\]