A number $ x$ is uniformly chosen on the interval $ [0,1]$, and $ y$ is uniformly randomly chosen on $ [\minus{}1,1]$. Find the probability that $ x>y$.
1. We need to find the probability that \( x > y \) where \( x \) is uniformly chosen from the interval \([0,1]\) and \( y \) is uniformly chosen from the interval \([-1,1]\).

2. First, consider the probability that \( y > x \). We will subtract this probability from 1 to get the desired probability.

3. The variable \( y \) can be in the interval \([0,1]\) with probability \(\frac{1-0}{1-(-1)} = \frac{1}{2}\). This is because the length of the interval \([0,1]\) is 1, and the length of the interval \([-1,1]\) is 2.

4. If \( y \) is in the interval \([0,1]\), we can visualize the problem in a unit square with vertices at \((0,0)\), \((0,1)\), \((1,0)\), and \((1,1)\).

5. The area where \( y > x \) within this unit square is the area above the line \( y = x \). This area forms a right triangle with legs of length 1.

6. The area of this right triangle is:
   \[
   \text{Area} = \frac{1 \cdot 1}{2} = \frac{1}{2}
   \]

7. Therefore, the probability that \( y > x \) when \( y \) is in \([0,1]\) is \(\frac{1}{2}\).

8. Since \( y \) is in \([0,1]\) with probability \(\frac{1}{2}\), the overall probability that \( y > x \) is:
   \[
   \text{Probability}(y > x) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
   \]

9. Finally, the probability that \( x > y \) is:
   \[
   \text{Probability}(x > y) = 1 - \text{Probability}(y > x) = 1 - \frac{1}{4} = \frac{3}{4}
   \]

The final answer is \(\boxed{\frac{3}{4}}\).