comparing two numbers and then finding out expectations

Question

Suppose X contains 2,3,5 and Y contains 4,5,6.Some number is randomly chosen from X and then another number from Y.What is the expected number of times that x>y?

Answer 1

So of the three choices you have in x, only one of them can result in an outcome in which $x > y$. If 5 is selected from $x$, then $x > y$ only if the 4 is selected from $y$. So, we have 3 possible selections from $x$, for each of those three there are 3 possible selections from $y$, leading to ${3 \choose 1} \cdot {3 \choose 1} = 3 \cdot 3 = 9$ possible outcomes. Of these 9 possible outcomes $x > y$ in 1 of them. So I suppose you could expect that out of 9 randomly pulls, $1$ would result in an outcome where $x > y$.

Answer 2

This can be approached in a rigorous way if you'd wish to say as well. Let's define the following function. We are given that $X,Y$ are independent Random Variables with uniform distributions and that $X$ takes on values $2,3,5$ and $Y$ takes on values $4,5,6$. Define the function $S=X+(-Y)$. Also define $Z=-Y$. Now we've defined two functions using which we can determine our required probability. The question asks us the probability that we'd get a number $X$ greater than $Y$ which means $P(\text{X>Y})=P(X-Y>0)=P(X+Z>0)=P(S>0)$ according to our definition.

By defining the above, we can determine the Probability mass function of $S$, $p_s(S)$ using the fact that it is the convolution of the PMFs $p_x(X)$ and $p_Z(Z)$ since $X$ and $Z$ are independent. If you perform the convolution, you'll see that the PMF of $S$ is: $$p_s(S=s)=\left\{\begin{array}{c}p_s(S=-4)=\frac{1}{9}\\p_s(S=-3)=\frac{2}{9}\\p_s(S=-2)=\frac{2}{9}\\p_s(S=-1)=\frac{2}{9}\\p_s(S=-0)=\frac{1}{9}\\p_s(S=1)=\frac{1}{9}\end{array}\right.$$ from which $P(S>0)=P(S=1)=p_s(S=1)=\frac{1}{9}$. Hence you'd expect on an average that out of $9$ turns, you get $1$ turn in which $X$ is greater than $Y$.

Answer 3

case 1: Let chosen number from bag X be 2

As numbers in bag Y are 4,5,6

Clearly 2 is smaller than 4,5,6

case 2: Let chosen number from bag X be 3

As numbers in bag Y are 4,5,6

Clearly 3 is smaller than 4,5,6

case 3: Let chosen number from bag X be 5

As numbers in bag Y are 4,5,6

Clearly $5 > 4$

Conclusion: Only one time $x>y$

Answer 4

Note that $$[X\gt Y]=[X=5]\cap[Y=4],$$ hence, by independence of $X$ and $Y$, $$P[X\gt Y]=P[X=5]\cdot P[Y=4]=\tfrac13\cdot\tfrac13.$$