Choosing Two Number from the Interval [0, 2] for Calculating P(A∩B) Question :

Alice and Bob each choose a number independently and uniformly at
random from the interval [0, 2]. Consider the following events :

*

*A : The absolute difference between the two numbers is greater than 1/4.

*B : Alice's number is greater than 1/4.

Compute Pr(A∩B).

My ans :
P(A)   = ( 2 - 1/4 ) / ( 2 ) = 0.875
P(B)   = ( 2 - 1/4 ) / ( 2 ) = 0.875
P(A∩B) =       0.875 * 0.875 = 0.765625

My friend obtains a different one = 0.866666...

I want to know the real answer of this question. This question is quite different for different people. What do you think? Please explain. =]
Thank you for you attenion.

[Update]
Answer found online : 85/128 = 0.6640625
Reference : http://www.scotthyoung.com/mit/6041-exam.pdf
 A: Outline: This can be viewed as a geometry problem. Draw the $2\times 2$ square with corners $(0,0)$, $(2,0)$, $(2,2)$, and $(0,2)$. Draw the lines $x-y=\frac{1}{4}$ and $x-y=-\frac{1}{4}$. Draw the line $x=\frac{1}{4}$.
We want the probability that the pair $(X,Y)$ lands in the part of the square which is not between the two lines, and is to the right of the line $x=\frac{1}{4}$.
To find that probability, (i) find the area $k$ of the part of the square which is not between the two lines and is to the right of $x=\frac{1}{4}$ and  (ii) divide $k$ by $4$.
Remark: You can calculate $\Pr(A\cap B)$ by a strategy of the kind you were attempting. The calculation of $\Pr(A)$ in the post is not right. The combined area of the two triangles that represent the event $A$ is $\left(2-\frac{1}{4}\right)^2$. Divide by $4$. We get $\Pr(A)=\frac{49}{64}$.
A: To be calculated is integral:
$\frac{1}{4}\int_{0}^{2}\int_{0}^{2}1_A\left(a,b\right)dadb$ where
$1_A\left(a,b\right)=1$ if $a>\frac{1}{4}\wedge\left|a-b\right|>\frac{1}{4}$
and $1_A\left(a,b\right)=0$ otherwise. 
Here $a$ corresponds with Alice and $b$ with Bob.
It comes to determining $\frac{1}{4}A$ where $A$ denotes area: $\left\{ \left(a,b\right)\in\left[0,2\right]^{2}\mid a>\frac{1}{4}\wedge\left|a-b\right|>\frac{1}{4}\right\} $.
