Joint distribution of X and Y bernoulli random variables A box contains two coins: a regular coin and a biased coin with $P(H)=\frac23$. I choose a coin at random and toss it once. I define the random variable X as a Bernoulli random variable associated with this coin toss, i.e., X=1 if the result of the coin toss is heads and X=0 otherwise. Then I take the remaining coin in the box and toss it once. I define the random variable Y as a Bernoulli random variable associated with the second coin toss.
a)Find the joint PMF of X and Y.
b)Are X and Y independent?
My attempt to answer this question:
Let A be the event that first coin, I pick is the regular(fair) coin. Then conditioning on that event, I can find joint PMF. Once conditioned, I can decide if X and Y are independent(conditionally).
$P(A)=\frac12,P(A^c)=\frac12$.
In the event A, $P(X=1)=\frac12,P(Y=1)=\frac23$.
In the event$A^c, P(X=1)=\frac23, P(Y=1)=\frac12$
So, $P_{X,Y}(x,y)= P(X=x, Y=y|A)P(A) + P(X=x, Y=y|A^c)P(A^c)$
$P_{X,Y}(x,y)=P_{\frac12}(x)P_{\frac23}(y)(\frac12) + P_{\frac23}(x)P_{\frac12}(y)(\frac12)$
Now, how can we find Joint PMF of X and Y using Bernoulli distribution?
 A: The joint pmf can be described by a 2-by-2 contingency table that shows the probabilities of getting $X=1$ and $Y=1$, $X=1$ and $Y=0$, $X=0$ and $Y=1$, $X=0$ and $Y=0$.
So you'll have:





$X=0$
$X=1$




$Y=0$
$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$
$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{4}$


$Y=1$
$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{4}$
$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$



A: Let $U$ take value $1$ if the regular coin results in heads and let $U$ take value $0$ otherwise.
Let $V$ take value $1$ if the biased coin results in heads and let $V$ take value $0$ otherwise.
Then by independence of $U$ and $V$:$$P(X=1,Y=1)=P(U=1,V=1)=P(U=1)P(V=1)=\frac12\frac23=\frac13$$
and:
$$P(X=0,Y=0)=P(U=0,V=0)=P(U=0)P(V=0)=\frac12\frac13=\frac16$$
Further by symmetry:$$P(X=1,Y=0)=P(X=0,Y=1)$$
so that we can conclude that:$$P(X=1,Y=0)=P(X=0,Y=1)=\frac12\left(1-\frac13-\frac16\right)=\frac14$$
