Question

Find \(P\left(0

Short answer

The joint probability \(P\left(0<X_{1}<\frac{1}{3}, 0<X_{2}<\frac{1}{3}\right)\) is approximately 0.066 or 6.6%

Step-by-step solution

Understand the Variables

In this problem, \(X_{1}\) and \(X_{2}\) are continuous random variables with a joint pdf equals to \(f\left(x_{1}, x_{2}\right)=4x_{1}\left(1-x_{2}\right)\). To find \(P\left(0<X_{1}<\frac{1}{3}, 0<X_{2}<\frac{1}{3}\right)\), we can use the double integral of the given joint pdf within the required range.

Set Up the Double Integral

Our double integral setup will look like this:\[\int_0^{1/3} \int_0^{1/3} 4 x_1 (1-x_2) dx_1 dx_2\]It can be recognized here that the order of integration does not matter since \(x_{1}\) and \(x_{2}\) are independent.

Calculate the Inner Integral

Evaluating the inner integral gives us:\[\int_0^{1/3}[2x_{1}^{2} - 4 x_{1}^{2}x_{2}]_0^{1/3} dx_2 = \int_0^{1/3} 2/9 - 4/27 x_{2} dx_2\]Now, we only have a simple integral in terms of \(x_{2}\) to solve.

Calculate the Outer Integral

Evaluating the outer integral gives:\[[2/9 x_{2} - 2/81 x_{2}^{2}]_0^{1/3} = 2/27 - 2/243 = 16/243\]This is the probability we were looking for. It means there's about 6.6% chance that the values of both \(X_{1}\) and \(X_{2}\) fall into the specified ranges.