We are given two real numbers. The first number, $x$, lies in the interval $[-2, 2]$. The second number, $y$, is positive and does not exceed 4, meaning $0 < y \le 4$. We need to find the probability that the second number $y$ is less than the square of the first number $x$, i.e., $y < x^2$.

1. Problem Description

We are given two real numbers. The first number, $x$, lies in the interval $[-2, 2]$. The second number, $y$, is positive and does not exceed 4, meaning $0 < y \le 4$. We need to find the probability that the second number $y$ is less than the square of the first number $x$, i.e., $y < x^2$.

2. Solution Steps

First, we define the region where the numbers $x$ and $y$ can exist.

Since $x$ belongs to $[-2, 2]$, we have $-2 \le x \le 2$.

Since $y$ is positive and does not exceed 4, we have $0 < y \le 4$.

The total region is a rectangle with vertices $(-2, 0), (2, 0), (2, 4), (-2, 4)$ in the $xy$-plane.

The area of this rectangle is $A_{total} = (2 - (-2)) \times (4 - 0) = 4 \times 4 = 16$.

Next, we need to find the area of the region where $y < x^2$.

The curve $y = x^2$ intersects the rectangle.

Since $0 < y \le 4$ and $-2 \le x \le 2$, we are interested in the area under the curve $y = x^2$ within the rectangle.

We need to integrate $y = x^2$ with respect to $x$ from $-2$ to $2$:

$A_{desired} = \int_{-2}^{2} x^2 dx$

$A_{desired} = [\frac{x^3}{3}]_{-2}^{2}$

$A_{desired} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$

The probability that $y < x^2$ is the ratio of the area under the curve $y = x^2$ to the total area of the rectangle:

$P(y < x^2) = \frac{A_{desired}}{A_{total}} = \frac{\frac{16}{3}}{16} = \frac{16}{3} \times \frac{1}{16} = \frac{1}{3}$

3. Final Answer

The probability is $\frac{1}{3}$.