To tackle this problem, let's first define and justify our choice of coordinate system, and then set up the necessary integrals for each part of the problem.

Coordinate System Justification

We are given a joint density function $f(x, y, z)$ over the intervals $[0, 1] \times [0, 1] \times [0, 1]$. Since the boundaries of the events are all within the unit cube, a Cartesian coordinate system $(x, y, z)$ is appropriate. This choice simplifies the integration process because the limits are clearly defined as $0 \leq x, y, z \leq 1$.

1. Verifying the Probability Density Function

To verify that $f(x, y, z)$ is a probability density function, we need to ensure that the total integral over the given domain equals 1:

[ f(x, y, z) = \begin{cases} 12x^2yz & \text{if } (x, y, z) \in [0, 1] \times [0, 1] \times [0, 1] \ 0 & \text{otherwise} \end{cases} ]

The integral should be set up as follows:

[ \iiint_{[0,1] \times [0,1] \times [0,1]} 12x^2yz , dx , dy , dz ]

2. Probability that $2x + y \leq 1$

To find the probability that a chosen point $(x, y, z)$ satisfies $2x + y \leq 1$, we need to integrate $f(x, y, z)$ over the region where $2x + y \leq 1$:

$\iiint_{D} 12x^2yz \, dx \, dy \, dz$

where $D$ is the region defined by $0 \leq x \leq 1$, $0 \leq y \leq 1 - 2x$, and $0 \leq z \leq 1$.

3. Expected Value of Each Variable

The expected value of a random variable $X$ with a joint density function $f(x, y, z)$ is given by:

[ E[X] = \iiint_{[0,1] \times [0,1] \times [0,1]} x \cdot f(x, y, z) , dx , dy , dz ]

Similarly, the expected values for $Y$ and $Z$ are:

[ E[Y] = \iiint_{[0,1] \times [0,1] \times [0,1]} y \cdot f(x, y, z) , dx , dy , dz ]

[ E[Z] = \iiint_{[0,1] \times [0,1] \times [0,1]} z \cdot f(x, y, z) , dx , dy , dz ]

Now, we can set up the specific integrals for each part.

Detailed Setup of Integrals

1. Verifying the Probability Density Function

$\iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} 12x^2yz \, dx \, dy \, dz$

2. Probability that $2x + y \leq 1$

$\iiint_{0}^{1} \iiint_{0}^{1 - 2x} \iiint_{0}^{1} 12x^2yz \, dz \, dy \, dx$

3. Expected Value of Each Variable

• $E[X]$:

$E[X] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} x \cdot 12x^2yz \, dx \, dy \, dz$

• $E[Y]$:

$E[Y] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} y \cdot 12x^2yz \, dx \, dy \, dz$

• $E[Z]$:

$E[Z] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} z \cdot 12x^2yz \, dx \, dy \, dz$

These integrals are correctly set up to determine the necessary probability and expected values based on the provided joint density function.

Do you want any details on the integration process or have any questions?

Here are five additional questions you might consider asking:

1. How do you perform these integrals step-by-step?
2. What are the interpretations of the expected values in probability?
3. How would the setup change if the limits of integration were different?
4. Can you explain how the region $2x + y \leq 1$ is determined geometrically?
5. How can these integrals be evaluated using numerical methods?

Tip: When dealing with joint density functions, visualizing the region of integration can greatly help in setting up the correct limits.