Calculate the probability that the product of two samples from X is greater than the sum of two samples from Y

Question

Suppose there is a distribution $X$ with $P(x) = 1/3, x = 1,2, 3 $ (0 otherwise) and another distribution $Y$ with distribution $P(y) = 1/y, y=2, 3, 6$ (0 otherwise).

Find for two independent samples from $X$ ($X_1, X_2$) and two independent samples from $Y$ ($Y_1, Y_2$) $P(X_1 X_2 > Y_1 + Y_2)$.

I know that this can be done by enumerating cases; is there an algebraic method to do so?

Answer 1

There are so few cases that it might easier just to enumerate all cases. In the process, you will find out that there are ways to extrapolate to more general cases with larger data set.

First, find the minimum of $Y_1+Y_2$, which is 4, so $X_1 X_2$ must be greater than 4, giving the only three possibilities: (2,3), (3,2), and (3,3). $X_1 X_2$ equals 6 and 9, respectively, and the probabilities are $P_1=2*(1/3)^2=2/9$ and $P_2=(1/3)^2=1/9$.

In the case of $X_1 X_2=6$, there are the following cases (listed in a general way)

• $Y_1=2$ ($P=1/2$), than $Y_2$ must be less than or equal to $3$, and $P(Y_2\le 3)=1/2+1/3=5/6$. Totally $P = 1/2 \times 5/6 = 5/12$.
• $Y_1=3$ ($P=1/3$), than $Y_2$ must be less than or equal to $2$, and $P = 1/2$. In total, $P = 1/3 \times 1/2 = 1/6$.

In the cases where $X_1 X_2 = 9$, we have

• $P(Y_1=2) \times P(Y_2 \le 6) = 1/2 \times 1 = 1/2$,
• $P(Y_1=3) \times P(Y_2 \le 5) = 1/3 \times (1/2 + 1/3) = 5/18$,
• $P(Y_1=6) \times P(Y_2 \le 2) = 1/6 \times 1/2 = 1/12$.

Finally, the answer to the question is $$ P = \frac{2}{9} \times \left( \frac{5}{12} + \frac{1}{6} \right) + \frac{1}{9} \times \left( \frac{1}{2} + \frac{5}{18} + \frac{1}{12} \right) = \frac{73}{324}. $$

Answer 2

Assuming choice without replacement. Only possibility $X_1X_2=6$ with probability $\frac{1}{3}$ and $Y_1+Y_2=5$ with probability $\frac{7}{12}$. Since events are independent. Desired $P=\frac{7}{36}$.