# Sum of two continuous random variables

Let R1 and R2 be two independent random variables, both with uniform density at the interval (0,2).

What is the probability of R1>1 given that R1 +R2<2?

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What I've tried:
I know that $$P(R1>1/R1+R2<2)= \frac{P(R1 +R2<2 ∩ R1>1)}{P(R1 +R2<2)}$$ And I know that P(R1>1)=1/2 and P(R1 +R2<2)=1/2

How can I find the intersection?

• You would need more information on the joint distribution of $X_1, X_2$ than that. Are they independent? If so, are they uniform? Nov 3, 2019 at 12:42
• Are they perhaps independent and identically distributed?
– whuber
Nov 3, 2019 at 13:38
• They are uniform and independent. I edited Nov 3, 2019 at 17:31

You just need to find $$p=P(R_1+R_2<2\cap R_1>1)$$. The joint distribution is $$1/4$$ in $$[0,2]\times [0,2]$$, and the probability of interest here is a triangular region between $$x+y=2$$,$$y=0$$,$$x=1$$, which has an area of $$1/2$$. So, $$p=(1/2)\times(1/4)=1/8$$.
$$P(R_1>1|R_1+R_2<2)=\frac{1/8}{1/2}=1/4$$
• @Oalvinegro Since they're independent, it is the multiplication of marginals, i.e. $1/2\times 1/2$ Nov 3, 2019 at 18:26