Score Probability Question (Checking my answer) I'm trying to check my answer to the follow:

John's score is a number uniformly chosen between $0$ and $3$. Mary's score is a number uniformly chosen between $0$ and $1$. The two numbers are chosen independently. Find the probability that John's score is at least twice Mary's score.

My attempt:
We want to find $P(J > 2M)$ where $J$ is the score John receives and $M$ is the score Mary receives.
I first found the CDF for Mary and John:
$$P(M \leq m) = m$$
$$P(J \leq j) = \frac{j}{3}$$
Then the probability that John's score is twice Mary's given we have Mary's score:
$$P(J > 2M \mid M) = \frac{3-2M}{3}$$
Using the total law of probability
$$P(J > 2M \mid M)P(M) = \frac{(3-2M)M}{3}$$ which should be the probability that John's score is at least twice as great as Mary's.
Is this correct?
 A: I'm assuming you're sure the question means they're continuous random variables (it reasonably could be interpreted as discrete if "between" could mean inclusive of the endpoints; in that case you'd just enumerate the events).  Then $(J,M)$ is uniformly distributed over the rectangle $[0,3]\times[0,1]$, and the probability of any set is just the area of the set divided by $3$ (the total area).  Then I suggest you sketch $\{(j,m): j\ge 2m\}$ to see that it's the union of a certain triangle and a $1\times1$ square, thus
$$P(J\ge 2M)=\frac{(1/2)\cdot 1\cdot 2+1}{3}=\frac{2}{3}.$$
A: You've correctly worked out $P(J>2M\mid M=m)=\frac{(3-2m)}{3}$. This is the probability for a particular value of Mary's score, $M=m$. The problem is that you don't know $M$, but only know its distribution. So your final expression doesn't make sense - it is still written in terms of an unknown variable whereas your answer should be a number.
If you had a discrete distribution for $M$ (I'll describe below how this changes for a continuous distribution, but it is easier to understand if you think about the discrete case first), you could now multiply by $P(M=m)$ to get $P(M=m,J>2M)$. But that would only count the successful cases where Mary gets one particular value $m$, and you would need to sum all these probabilities for the different $m$ to find the total probability.
You have a continuous distribution, so what you actually need to do at this point is average $P(J>2M\mid M=m)$ over the distribution of Mary's score. That is you want $\int_0^1 \frac{(3-2m)}{3}f(m)\,dm$, where $f(m)$ is the pdf of Mary's score (here that is just $1$ on the range of possible values). This is exactly the continuous analogue of the discrete sum above.
