If one number from each set is selected at random, then determine $P(Y < X)$

Question

A discrete random variable, $X$, is uniformly distributed between $1$ and $10$ inclusive, and another discrete random variable, $Y$, is uniformly distributed between $5$ and $15$ inclusive. If one number from each set is selected at random, then determine $P(Y < X)$.

I understand that this would mean $$P(X=1)=\frac{1}{10},\;P(X=2)=\frac{1}{10},\;\ldots,\;P(X=10)=\frac{1}{10},$$ and $$P(Y=5)=\frac{1}{10},\;P(Y=6)=\frac{1}{10},\;\ldots,\;P(Y=15)=\frac{1}{10}.$$

How should I start this and answer the question?

Answer 1

$$P(Y < X) = \bigcup_i P(Y < X, Y = i) = \bigcup_i P(Y<X | Y = i) P(Y = i) = \sum_{i=5}^{9} \left(1 - \frac{i}{10} \right)\frac{1}{11} = \frac{3}{22}. $$

EDIT: As was pointed out, I had a typo and the probability $P(Y = i) = 1/11$ and not $1/16$.

Answer 2

As mentioned in the comments we need to start with $P(X=1)=\cdots=P(X=10)=\frac{1}{10}$ and $P(Y=5)=\cdots=P(Y=15)=\frac{1}{11}$. The answer given by Gregory has the essential elements but was written hastily I suspect and has some errors;) Here is a bit more detail, and hopefully error free:

\begin{eqnarray*} P(Y<X) &=& P\Big(\bigcup_{j=5}^{15}(Y<X,Y=j)\Big)\ \text{(reason: partitioning }(Y<X)\text{ over disjoint sets }(Y=j)\text{)}\\ && \sum_{j=5}^{15}P(Y<X,Y=j)\ \text{(reason: probability of union of disjoint sets)}\\ &=& \sum_{j=5}^{15}P(j<X,Y=j)\ \text{(reason: }Y=j\text{ on each set)}\\ &=& \sum_{j=5}^{9}P(j<X,Y=j)\ \text{(reason: }X\text{ can only be 1-10)}\\ &=& \sum_{j=5}^{9}P(j<X)P(Y=j)\ \text{(reason: }X,Y\text{ are independent)}\\ &=& \sum_{j=5}^{9}\Big(1-P(X\leq j)\Big)P(Y=j)\ \text{(reason: }P(B)=1-P(B^c)\text{)}\\ &=& \sum_{j=5}^{9}\Big(1-\frac{j}{10}\Big)P(Y=j)\ \text{(reason: }P(X\leq j)= \frac{1}{10}+\cdots+\frac{1}{10},\ j\text{ times)}\\ &=& \sum_{j=5}^{9}\Big(1-\frac{j}{10}\Big)\frac{1}{11}\ \text{(reason: }P(Y=j)=\frac{1}{11}\text{)}\\ &=& \Big(5-\frac{35}{10}\Big)\frac{1}{11} = \frac{15}{110} = \frac{3}{22} \end{eqnarray*}