How to compute $\mathbb{P}(A|B)$ of two independent RVs? There are two independent RVs $X \sim \mathcal{U}(-1,5)$ and $Y \sim \mathcal U(-5,5)$.
Let $A = \{ X \ge Y \land Y \ge -1 \land Y \le 1\}$ and $B = \{ X \le 1\}$.
What is $\mathbb P (A | B)$?
My idea is to apply Bayes' rule, i.e.
$$ \mathbb P(A|B) = \frac{\mathbb P(B|A) \mathbb P(A)}{\mathbb P(B|A)\mathbb P(A) + \mathbb P(B|\lnot A) \mathbb P(\lnot A)}$$
I understand that this works nicely as long as $\mathbb P(B|A)$ is easier to compute.
However, I find that
$$ \mathbb P(B|A) = \frac{\mathbb P(A|B) \mathbb P(B)}{\mathbb P(A|B)\mathbb P(B) + \mathbb P(A|\lnot B) \mathbb P(\lnot B)}$$
... which I feel does not bring me closer to the solution. The unconditional probabilities are fairly straight forward, but the conditional ones keep piling up.
How should I procede from there?
 A: To follow your approach, note the definition of conditional probability $p(A|B) =\frac{p(A \land B)}{p(B)}$. Knowing this, one just need solve for $p(A \land B)$, as calculating $p(B)$ is straight-forward.
We're now looking for $A\land B = \{ X \ge Y \land Y \ge -1 \land Y \le 1 \land X \le 1\}$. Written to stress the ordering of values gives 
$$A\land B = \{ Y \le X \land X \le 1 \land -1 \le Y \land Y \le 1 \}$$
Again by conditional probability:
$$p(A\land B) = p(\{ Y \le X | X \le 1 \land -1 \le Y \land Y \le 1 \}) p(\{X \le 1 \land -1 \le Y \land Y \le 1 \})$$
Since $X, Y$ are independent, the second term may be factored into the product $p(X\le 1)p(-1 \le Y  \le 1)$, and each factor can be calculated from its respective distribution. Solve for the first factor:
$$p(\{ Y \le X | X \le 1 \land -1 \le Y \land Y \le 1 \}) = \int_{-1}^{1}\int_{-1}^{x}\frac{dxdy}{4} = \frac{1}{4}\int_{-1}^1(x+1)dx = \frac{1}{2}$$
Combining all factors:
$$p(A|B) = \frac{p(A\land B)}{p(B)} $$
$$= \frac{p(\{ Y \le X | X \le 1 \land -1 \le Y \land Y \le 1 \}) p(X\le 1)p(-1 \le Y  \le 1)}{p(B)}$$
$$\frac{1}{2}\frac{1}{5} = \frac{1}{10}$$
More intuitively: Knowing that $X \le 1$ turns the sample space into bivariate uniform on $x \in [-1,1], y \in [-5, 5]$; the total area of this space is $20$, and the area of the triangle bound by $y = -1, x = 1, y = x$ is $2$. An integral based on this approach gives an equivalent answer:
$$\int_{-1}^1\int_{-1}^{x}\frac{dx dy}{20} = \frac{1}{20}\int_{-1}^1(x+1)dx = \frac{1}{20}2= \frac{1}{10}$$
