Joint probability mass function $f$ of $2$ random variables.

Joint probability mass function $f$ of $2$ random variables.

Given $\Omega$={$20, 21, ..., 49$}, $P$({$i$})=\begin{cases}0.05, &
\text{if $20<=i<=29$} \\0.03, & \text{if $30<=i<=39$} \\0.02, & \text{if
$40<=i<=49$}\end{cases}
The random variables $X, Y$ are defined as:
$X$($i$)=\begin{cases}5, & \text{if $20<=i<=24$} \\10, & \text{if
$25<=i<=34$} \\15, & \text{if $35<=i<=49$}\end{cases}
$Y$($i$)=\begin{cases}10, & \text{if $20<=i<=23$} \\5, & \text{if
$24<=i<=28$} \\15, & \text{if $29<=i<=39$} \\10, & \text{if
$40<=i<=49$}\end{cases}
Do $X$ and $Y$ have the same distribution? Are $X$ and $Y$ indepedent?
Calculate the joint probability mass function $f$ of $(X,Y)$ for:
$f(5,5)$, $f(5,10)$ and $f(5,15)$.
Ok here my solutions:
$X$ and $Y$ do not have the same distribution since: $P(X=10)=P(25)=0.05$,
$P(Y=10)=P(40)=0.02$.
$X$ and $Y$ are not indepedent since:
$P(X=5)*P(Y=5)=P(20)*P(24)=0.05*0.05=0.025$. $P(X=5, Y=5)=P(24)=0.05$
$f(5,5)=P(X=5, Y=5)=P(24)=0.05$
$f(5,10)=P(X=5, Y=10)=P(20)=0.05$
$f(5,5)=P(X=5, Y=15)=0$
Is everything correct?