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Let X be a continuous uniform R.V. between $x_a$ and $x_b$ and $Y_1$ and $Y_2$ be identically distributed continuous uniform R.V. between $y_a$ and $y_b$. $y_a$ and $y_b$ and sandwiched in between $x_a$ and $x_b$. All three R.V.'s are independent. What's the probability that X is realized smaller than the two Y's?

We take the probability that X is realized from $x_a$ to $y_a$ and add it to the probability that X is realized within $y_a$ and $y_b$ AND is less than both $Y_1$ and $Y_2$. I had previously asked how one can determine that one R.V. is realized smaller than another random variable, but the question of whether it's smaller than two (or potentially even more R.V.'s) has me puzzled.

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$\Pr(x_a \le X \le y_a) = \dfrac{y_a-x_a}{x_b-x_a}$ and $\Pr(X \lt \min(Y_1,Y_2)|x_a \le X \le y_a)=1$;

$\Pr(y_a \le X \le y_b) = \dfrac{y_b-y_a}{x_b-x_a}$ and $\Pr(X \lt \min(Y_1,Y_2)|y_a \le X \le y_b)=\dfrac13$;

$\Pr(y_b \le X \le x_b) = \dfrac{x_b-y_b}{x_b-x_a}$ and $\Pr(X \lt \min(Y_1,Y_2)|y_b \le X \le x_b)=0$;

so $\Pr(X \lt \min(Y_1,Y_2)) = \dfrac{y_a-x_a}{x_b-x_a} + \dfrac13 \left(\dfrac{y_b-y_a}{x_b-x_a}\right) = \dfrac{y_b+2y_a-3x_a}{3(x_b-x_a)}.$

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Hint: you can write $$P(X \le Y_1, X \le Y_2) = P(X \le Y_1 \le Y_2) + P(X \le Y_2 \le Y_1).$$ (Draw a Venn diagram if necessary.) The two probabilities on the right side are equal (why?) so you only need to compute one of them. Now note that $P(X \le Y_1 \le Y_2)$ is really just computing the volume of a certain subset of the box $[x_a, x_b] \times [y_a, y_b] \times [y_a, y_b]$ which could be an easy calculus exercise.

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That box is a nice way to think about it, but I can't visualize X. I know that Y_1 less than Y_2 looks like a triangle (subset of a square), but then I don't know how it looks like in the X direction on top of that triangle. Does this subset looks like a 'piece of cake' (literally, not figuratively'() – Wuschelbeutel Kartoffelhuhn Sep 20 '12 at 4:21

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