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 analyze the events A and B defined by the inequalities involving the points (x, y) in the unit square [0, 1] x [0, 1].

### Step 1: Define the Events
- Event A: The points (x, y) satisfy \( y^2 \geq x \).
- Event B: The points (x, y) satisfy \( x^2 \geq y \).

### Step 2: Graph the Events
To visualize these events, we can graph the curves defined by the inequalities:
- The curve for event A is given by \( y = \sqrt{x} \).
- The curve for event B is given by \( y = x^2 \).

### Step 3: Find the Area for Event A
To find the probability of event A, we need to calculate the area under the curve \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 1 \).

\[ P(A) = \int_0^1 \sqrt{x} \, dx \]

Calculating the integral:

\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \bigg|_0^1 = \frac{2}{3}(1) - \frac{2}{3}(0) = \frac{2}{3} \]

Thus, \( P(A) = \frac{2}{3} \).

### Step 4: Find the Area for Event B
Next, we find the probability of event B by calculating the area under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \).

\[ P(B) = \int_0^1 x^2 \, dx \]

Calculating the integral:

\[ \int x^2 \, dx = \frac{1}{3} x^3 \bigg|_0^1 = \frac{1}{3}(1) - \frac{1}{3}(0) = \frac{1}{3} \]

Thus, \( P(B) = \frac{1}{3} \).

### Step 5: Find the Area for Event A Intersection B
To find the probability of the intersection of events A and B, we need to find the area that satisfies both inequalities. This area is bounded by the curves \( y = \sqrt{x} \) and \( y = x^2 \).

To find the points of intersection, set \( \sqrt{x} = x^2 \):

\[ x^{1/2} = x^2 \implies x^{1/2} - x^2 = 0 \implies x^{1/2}(1 - x^{3/2}) = 0 \]

This gives \( x = 0 \) and \( x = 1 \).

Now, we need to find the area between these curves from \( x = 0 \) to \( x = 1 \):

\[ P(A \cap B) = \int_0^1 (\sqrt{x} - x^2) \, dx \]

Calculating the integral:

\[ \int (\sqrt{x} - x^2) \, dx = \int \sqrt{x} \, dx - \int x^2 \, dx = \frac{2}{3} x^{3/2} \bigg|_0^1 - \frac{1}{3} x^3 \bigg|_0^1 \]

Calculating both parts:

\[ = \left(\frac{2}{3}(1) - 0\right) - \left(\frac{1}{3}(1) - 0\right) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

Thus, \( P(A \cap B) = \frac{1}{3} \).

### Step 6: Analyze the Results
Now we have:
- \( P(A) = \frac{2}{3} \)
- \( P(B) = \frac{1}{3} \)
- \( P(A \cap B) = \frac{1}{3} \)

### Step 7: Check the Options
1. \( P(A \cap B) = \frac{1}{3} \) (True)
2. \( A \subseteq B \) (False)
3. \( 2P(A) = 3P(B) \) (False, since \( 2 \times \frac{2}{3} \neq 3 \times \frac{1}{3} \))
4. \( P(B) < P(A) \) (True)

Thus, the correct answer is that \( P(A \cap B) = \frac{1}{3} \).