The discrete random variables X and Y are jointly distributed with joint mass function given in the table below:

$$ \begin{array}{|c|c|c|c|c|} \hline y / x & 0 & 3 & 6 \\ \hline 1& a & b & c \\ \hline 2 & 0.10 & 0.05 & d \\ \hline \end{array} $$

It is also known that $P(Y = 2|X = 0) = 0.25$ and that X and Y are independent. Four values in this table have been replaced by the letters a, b, c and d but we have enough information to calculate them.

What is the value of c (i.e. the value of $P(X = 6, Y = 1)$)?


$P(Y=2 | X = 0) = \frac{P(X=0 \cap Y=2)}{P(X=0)} = 0.25$

Since X and Y are indepedent. $P(Y = 2) = 0.25$

Not sure how to go about this.

  • $\begingroup$ Good start. So you got the second row sum to be $0.25$ and you can compute $d$. Using complementary property you can compute $\Pr\{X = 1\}$, and hence know that each unknown is $3$ times of the number in the second row. $\endgroup$ – BGM Aug 8 '18 at 5:07

How do you find $P(Y=2)$ based on the table?

$P(Y=2) = P(Y=2, X=0) + P(Y=2, X=3) + P(Y=2, X=6)$. This allows you to solve for $d$.

By independence, $P(X=6 \mid Y=1) = P(X=6 \mid Y=2)$. Can you see how this allows you to solve for $c$?

$\frac{P(X=6, Y=1)}{P(Y=1)} = \frac{P(X=6, Y=2)}{P(Y=2)}$. You know $P(Y=1)$ and $P(Y=2)$ from the previous computation. Moreover $P(X=6, Y=2) = d$ which you already computed, and $P(X=6, Y=1)=c$.


Because $X,Y$ are independent, $P(X=0\cap Y=2)=P(X=0)P(Y=2)\rightarrow P(Y=2)=0.25$

$P(Y=1)=0.25,P(Y=2)=0.75 \rightarrow P(Y=2\cap X=0)=P(Y=2)P(X=0)=0.1,P(X=0)=0.1\times4=0.4.$

Similarily,$P(X=3)=\frac{P(Y=2\cap X=3)}{P(Y=2)}=\frac{0.05}{0.25}=0.2{5}\rightarrow P(X=6)=1-0.4-0.2=0.4$

$c=P(X=6\cap Y=1)=P(X=6)P(Y=1)=0.4\times 0.75=0.3$

And all of the numbers on the table could be calculated easily.


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