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Given X~unif(a, b) and Y~unif(c, d) with a < c < b < d.

What's the probability that Y>X and Y being realized in the interval (c, b)?

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    @WuschelbeutelKartoffelhuhn: The right triangle has "legs" along $x=b$ and $y=c$. The top right-hand corner of the triangle is at $(b,b)$, bottom left is at $(c,c)$. We subtract the aarea of this from the area of a certain rectangle. Or else can find area of shape by using formula for area of a trapezoid.2012-08-30

2 Answers 2

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Here we assume that $X$ and $Y$ are independent random variables with uniform distributions as specified by OP.

Write out the definition of the probability: $ \mathsf{Pr}\left(Y>X, c Can you finish it off?

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    Sorry, I disambiguated the expression, and put in parenthesis.2012-08-30
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Here is one solution without explicit integration. Let $A$ be the event that $Y>X$ and $B$ the event that $c. Then we know that $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$. The two probabilities on the right are simpler to calculate. First $\mathrm{Pr}(B)=\frac{b-c}{d-c}$.

Next we get a handle on $\mathrm{Pr}(A|B)$. Split the event $A$ into two disjoint events: $A_1$ and $A_2$ where $A_1$ is the event that $Y>X$ and $X; and $A_2$ is the event that $Y>X$ and $X>c$. Then $\mathrm{Pr}(A_1)=\mathrm{Pr}(X and $\mathrm{Pr}(A_2)=\frac{1}{2}\mathrm{Pr}(X>c)=\frac{1}{2}\left(\frac{b-c}{b-a}\right)$. Putting these two together, we have that $\mathrm{Pr}(A|B)=\frac{c-a}{b-a}+\frac{1}{2}\left(\frac{b-c}{b-a}\right)$.

Finally, using $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$ and doing some algebra we find that $\mathrm{Pr}(A \cap B)=\frac{-2ab+b^2+2ac-c^2}{2(a-b)(c-d)}$.

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    Two things are unclear for me here: In Pr(A_2) where did the factor 1/2 come from and how was Y>X taken into account? Maybe the question answers itself, but I don't see how.2012-09-02