If $X$ and $Y$ are both independent and uniformly distributed on $[0, 1]$ . Define $h$ as the density of the random variable $Z = X +Y$ . What is $$\Pr\left(X ≤ \frac{1}{2}|X + Y ≥ \frac{5}{4}\right)?$$ I am able to find
$$h(z) =\begin{cases}z &:&z\text{ from }0\text{ to }1\\2-z&:&z\text{ from }1\text{ to }2\end{cases}$$
But I am unable to find the $$\Pr\left(X ≤ \frac{1}{2}|X + Y ≥ \frac{5}{4}\right).$$ Please help me
To derive $P(X\leq\frac{1}{2}|X+Y\geq\frac{5}{4})$, we need to compute $P(X+Y\geq\frac{5}{4})$ and $P(X\leq\frac{1}{2},X+Y\geq\frac{5}{4})$. From the independence of $X,Y$ we can easily derive:$f_{X,Y}(x,y)=f_X(x)f_Y(y)=1$.
Then:(You can do these integrals by drawing areas,without difficult calculation)
$$P(X+Y\geq\frac{5}{4})=\int_{x+y\geq\frac{5}{4},x\in[0,1],y\in[0,1]}f_{X,Y}(x,y)dxdy=\frac{9}{32}$$
$$P(X\leq\frac{1}{2},X+Y\geq\frac{5}{4})=\int_{x+y\geq\frac{5}{4},x\in[0,1],y\in[0,1],x\leq\frac{1}{2}}f_{X,Y}(x,y)dxdy=\frac{1}{32}$$
Therefore:
$$P(X\leq\frac{1}{2}|X+Y\geq\frac{5}{4})=\frac{P(X\leq\frac{1}{2},X+Y\geq\frac{5}{4})}{P(X+Y\geq\frac{5}{4})}=\frac{1}{9}$$