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

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?

• Please show your attempt. $X_1 X_2 = \frac{1}{9}$ so there are not too many cases you are looking at. May 6, 2021 at 16:06
• Looking at two answers, it looks like there is an ambiguity in the statement. Is it possible to choose the same $X$ or the same $Y$ twice? May 6, 2021 at 16:46
• The samples are independent (without replacement). Updated question to reflect this. May 6, 2021 at 18:37
• Enumeration question: without replacement, there is only one case! May 6, 2021 at 23:55

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 $$P = \frac{2}{9} \times \left( \frac{5}{12} + \frac{1}{6} \right) + \frac{1}{9} \times \left( \frac{1}{2} + \frac{5}{18} + \frac{1}{12} \right) = \frac{73}{324}.$$

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 $$P=\frac{7}{36}$$.