Let X and Y be independent random variables, uniform on [0,1]. If the smaller is less that 1/4, Find the conditional probability that larger is greater than 3/4.
My solution:
$P(\text{Required})=P(X>3/4|X>Y)P(X>Y)+P(Y>3/4|Y>X)$
Now
$P(X>3/4|X>Y)P(X>Y)=P(X>3/4,X>Y)=P(X>3/4, Y<1/4)=P(X>3/4)P(Y<1/4)=(1-3/4)(1/4-0)=1/16$
Similarly, $P(Y>3/4|Y>X) = 1/16$
Hence, $P(\text{Required})=1/8$
Is this correct?
Hint:
$\Pr\left(\max\left(X,Y\right)>\frac{3}{4}\mid\min\left(X,Y\right)<\frac{1}{4}\right)\Pr\left(\min\left(X,Y\right)<\frac{1}{4}\right)=\Pr\left(\max\left(X,Y\right)>\frac{3}{4}\wedge\min\left(X,Y\right)<\frac{1}{4}\right)$
$\Pr\left(\max\left(X,Y\right)>\frac{3}{4}\wedge\min\left(X,Y\right)<\frac{1}{4}\right)=\Pr\left(X>\frac{3}{4}\wedge Y<\frac{1}{4}\right)+\Pr\left(Y>\frac{3}{4}\wedge X<\frac{1}{4}\right)$
$\Pr\left(\min\left(X,Y\right)<\frac{1}{4}\right)=1-\Pr\left(\min\left(X,Y\right)>\frac{1}{4}\right)=1-\Pr\left(X>\frac{1}{4}\wedge Y>\frac{1}{4}\right)$
You will end up with $\frac27$.
Split up the square $[0,1]\times[0,1]$ in $4\times4=16$ smaller squares. Condition $\min(X,Y)<\frac14$ demands to focus on $7$ of these smaller squares and in $2$ of them it is true that $\max(X,Y)>\frac34$.