If x and y co-ordinates of any point P are chosen randomly from intervals `[0, 2]` and `[0, 1]` respectively, then the probability `y<=x^2` is (A) `1/2` (B) `2/3` (C) `3/4` (D) `1/4`

Options

A. `1/2`
B. `2/3`
C. `3/4`
D. `1/4`

Answer

To solve the problem of finding the probability that \( y \leq x^2 \) when \( x \) is chosen randomly from the interval \([0, 2]\) and \( y \) is chosen randomly from the interval \([0, 1]\), we can follow these steps:

### Step 1: Understand the regions
We need to analyze the area defined by the conditions \( x \in [0, 2] \) and \( y \in [0, 1] \). The equation \( y = x^2 \) describes a parabola that opens upwards.

### Step 2: Identify the area of interest
The area of interest is the region where \( y \leq x^2 \) within the rectangle defined by the intervals for \( x \) and \( y \). The rectangle has vertices at \( (0, 0) \), \( (2, 0) \), \( (2, 1) \), and \( (0, 1) \).

### Step 3: Calculate the area of the rectangle
The area \( A_R \) of the rectangle can be calculated as:
\[ A_R = \text{length} \times \text{breadth} = 2 \times 1 = 2 \]

### Step 4: Calculate the area under the curve \( y = x^2 \)
We need to find the area under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \) (since \( y \) can only go up to 1). The area under the curve can be calculated using integration:
\[ \text{Area under } y = x^2 = \int_0^1 x^2 \, dx \]
Calculating the integral:
\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Step 5: Add the area of the rectangle above the curve
The area above the curve \( y = x^2 \) and below \( y = 1 \) from \( x = 0 \) to \( x = 1 \) is:
\[ \text{Area of rectangle above the curve} = \text{base} \times \text{height} = 1 \times (1 - \frac{1}{3}) = 1 \times \frac{2}{3} = \frac{2}{3} \]

### Step 6: Total area of the shaded region
The total area of the shaded region (where \( y \leq x^2 \)) is the area under the curve plus the area of the rectangle from \( x = 1 \) to \( x = 2 \):
\[ \text{Area of shaded region} = \text{Area under } y = x^2 + \text{Area of rectangle from } x = 1 \text{ to } x = 2 \]
The area from \( x = 1 \) to \( x = 2 \) is:
\[ \text{Area} = \text{base} \times \text{height} = 1 \times 1 = 1 \]
Thus, the total area of the shaded region is:
\[ \text{Area of shaded region} = \frac{1}{3} + 1 = \frac{4}{3} \]

### Step 7: Calculate the probability
The probability \( P \) that \( y \leq x^2 \) is given by the ratio of the area of the shaded region to the area of the rectangle:
\[ P = \frac{\text{Area of shaded region}}{\text{Area of rectangle}} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3} \]

### Final Answer
The required probability that \( y \leq x^2 \) is \( \frac{2}{3} \).