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I have this probability table (function of $(x,y)$)

$$\begin{array}{|c|ccc|} \hline X \setminus Y & 1 & 2 & 3 \\\hline 1 & a & b & c \\ 2 & 0.08 & 0.07 & d \\ \hline \end{array}$$

I have to find $a,b,c$ and $d$. I know that $X$, $Y$ are independent and that $P(X=2|Y=1)=0.3$.

Well, I found $d$ relatively easily... Since they are independent I found that $P(X=2) = 0.3$, therefore $d=0.15$.

But could anyone help me find $a, b$ and $c$? I am trying somehow to reclaim this expression $p(x,y) = p(x)\cdot p(y)$ but I can't. Actually I only need to find $a$ and $b$, I can get $c$ easily since the sum of the table is $1$. But how?

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1 Answer 1

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We use the independence to calculate $a$, $b$, and $c$. The three calculations are virtually identical, but the calculation of $c$ is numerically simpler, so we start with that.

By your calculation, we have $P(X=2,Y=3)=0.15$. By independence we have $$P(X=2,Y=3)=P(X=2)P(Y=3).$$ From the table, we have $P(X=2)=0.3$ and $P(Y=3=c+0.15$ (the sum of the entries in the third column). Thus we obtain the equation $$0.15=(0.3)(c+0.15).$$ Expand, solve the linear equation. We get $c=0.35$.

The values of $b$ and $a$ are found in exactly the same way. But because the answers are a little messier, we do the calculation. We have, from the table, $P(X=2,Y=2)=0.07$. But $P(X=2,Y=2)=P(X=2)P(Y=2)=(0.3)(b+0.07)$. So we need to solve the equation $0.07=(0.3)(b+0.07)$. We find that $b=\frac{(0.07)(0.7)}{0.3}$. This simplifies to $\frac{49}{300}$.

A similar calculation yields $a=\frac{(0.08)(0.7)}{0.3}=\frac{56}{300}$.

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Hmm , i see, i was very close i guess... BUT curiously enough, the total sum is not 1 , but 0.97666 ... –  Jans S Jun 2 '12 at 13:39
I had a typo for $a$. The sum is $1$. Bottom row has sum $0.3$. Top row has sum (backwards) $0.35+\frac{49}{300}+\frac{56}{300}=0.7$. –  André Nicolas Jun 2 '12 at 13:41
ah i see, kudos mate :-) –  Jans S Jun 2 '12 at 13:42

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