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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?

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

1 Answer 1

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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, $p=(1/2)\times(1/4)=1/8$.

$$P(R_1>1|R_1+R_2<2)=\frac{1/8}{1/2}=1/4$$

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  • $\begingroup$ How do I find the joint distribution? $\endgroup$
    – Oalvinegro
    Nov 3, 2019 at 18:25
  • $\begingroup$ @Oalvinegro Since they're independent, it is the multiplication of marginals, i.e. $1/2\times 1/2$ $\endgroup$
    – gunes
    Nov 3, 2019 at 18:26

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