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Consider three independent uniformly distributed (taking values between 0 and 1) random variables. What is the probability that the middle of the three values (between the lowest and the highest value) lies between a and b where 0 ≤ a < b ≤ 1 ?

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What did you try to solve this? –  Did Apr 19 '13 at 9:49
    
tried to find the range of the left point and that of the right point so that their mid-point lies within a and b. –  user2204800 Apr 19 '13 at 12:43

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Our probability space $\Omega$ is the unit cube $[0,1]^3$, and fate chooses a point $\omega=(x,y,z)\in\Omega$. That the three quantities $x$, $y$, $z\ $ "are independent and uniformly distributed" in $\Omega$ means the following: For any reasonable set $S\subset\Omega$ the probability that the point $\omega$ chosen by fate lies in $S$ is just the volume of $S$. (Note that ${\rm vol}(\Omega)=1$ in this example.)

We now consider a particular $S\subset\Omega$ as follows: $$ S:=\{(x,y,z)\>|\>x\leq y\leq z\ \ \wedge\ \ a\leq y\leq b\}\ .$$ Since the probability $P$ that the chosen point $\omega$ lies in $S$ is equal to ${\rm vol}(S)$ this probability can be written as a triple integral with $y$ as outmost variable: $$P={\rm vol}(S)=\int_a^b \Bigl(\int_0^y dx\Bigr)\Bigl(\int_y^1 dz\Bigr)\ dy=\int_a^b y(1-y)\ dy\ .$$ Now the six orderings of $x$, $y$, $z$ are equiprobable and mutually exclusive. Therefore the probability you are interested in is $\ 6P$.

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Dear @Christian, thanks a lot for the answer. :) This is for my own understanding. You are taking y as the point in between x and y, where 0 <= x <= y and y <= z <= 1 and a <= y <= b. But how does this ensure that y is the mid-point on the way from x to z? Also could you please suggest any book/study material for developing a better understanding to solve this kind of problems. –  user2204800 Apr 19 '13 at 15:11

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