Given that `x in [0,1]` and `y in [0,1]`. Let `A` be the event of selecting a point `(x,y)` satisfying `y^(2) ge x` and `B` be the event selecting a point `(x,y)` satisfying `x^(2) ge y`, then

Options

A. `P(AnnB)=(1)/(3)`
B. `AsubB`
C. `2P(A)=3P(B)`
D. `P(B) lt P(A)`

Explanation

To solve the problem, we need to find the probabilities of the events A and B, and then the intersection of these two events.

### Step 1: Define the Events

- **Event A**: Selecting a point \((x, y)\) such that \(y^2 \geq x\).
- **Event B**: Selecting a point \((x, y)\) such that \(x^2 \geq y\).

### Step 2: Find the Area for Event A

1. The inequality \(y^2 \geq x\) can be rewritten as \(y \geq \sqrt{x}\) (considering only the positive root since \(y\) is in \([0, 1]\)).
2. We need to find the area under the curve \(y = \sqrt{x}\) from \(x = 0\) to \(x = 1\).

The area can be calculated using integration:
\[ P(A) = \int_{0}^{1} \sqrt{x} \, dx \]

3. To compute the integral:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Therefore,
\[ P(A) = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3} \]

### Step 3: Find the Area for Event B

1. The inequality \(x^2 \geq y\) can be rewritten as \(y \leq x^2\).
2. We need to find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 1\).

The area can be calculated using integration:
\[ P(B) = \int_{0}^{1} x^2 \, dx \]

3. To compute the integral:
\[ \int x^2 \, dx = \frac{1}{3} x^3 \]
Therefore,
\[ P(B) = \left[ \frac{1}{3} x^3 \right]_{0}^{1} = \frac{1}{3} (1^3 - 0^3) = \frac{1}{3} \]

### Step 4: Find the Area of Intersection \(P(A \cap B)\)

1. The area of intersection is where both conditions \(y^2 \geq x\) and \(x^2 \geq y\) hold.
2. We need to find the area between the curves \(y = \sqrt{x}\) and \(y = x^2\).

3. First, find the points of intersection:
\[ \sqrt{x} = x^2 \implies x^{1/2} = x^2 \implies x^{1/2 - 2} = 1 \implies x^{-3/2} = 1 \implies x = 1 \]
The other point is \(x = 0\).

4. Now, we calculate the area between the curves from \(x = 0\) to \(x = 1\):
\[ P(A \cap B) = \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]

5. Compute the integral:
\[ \int (\sqrt{x} - x^2) \, dx = \int \sqrt{x} \, dx - \int x^2 \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} - \left[ \frac{1}{3} x^3 \right]_{0}^{1} \]
\[ = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Final Results

- \(P(A) = \frac{2}{3}\)
- \(P(B) = \frac{1}{3}\)
- \(P(A \cap B) = \frac{1}{3}\)

### Summary of the Solution

The probabilities for the events are:
- \(P(A) = \frac{2}{3}\)
- \(P(B) = \frac{1}{3}\)
- \(P(A \cap B) = \frac{1}{3}\)