Given the problem: a joint density function for probability of three events is
given by the following equation:
f (x, y, z) ={(12x2yz  on [0, 1] × [0, 1] × [0, 1])(0 otherwise)

and the Previous question of: What is the probability that a chosen point (x, y, z) is such that 2x + y ≤ 1

Answer:
What is the expected value of each variable?

To find the expected value of each variable \( X \), \( Y \), and \( Z \) given the joint density function, we use the definition of the expected value for continuous random variables. Given the joint density function \( f(x, y, z) \):

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

We use the expected value formulas:

\[ E[X] = \iiint x f(x, y, z) \, dx \, dy \, dz \]
\[ E[Y] = \iiint y f(x, y, z) \, dx \, dy \, dz \]
\[ E[Z] = \iiint z f(x, y, z) \, dx \, dy \, dz \]

### Expected Value of \( X \)

\[ E[X] = \iiint_0^1 x \cdot 12x^2yz \, dx \, dy \, dz \]

Separate the integrals:

\[ E[X] = 12 \int_0^1 \int_0^1 \int_0^1 x^3 yz \, dx \, dy \, dz \]
\[ E[X] = 12 \left( \int_0^1 x^3 \, dx \right) \left( \int_0^1 y \, dy \right) \left( \int_0^1 z \, dz \right) \]

Compute each integral separately:

\[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \]
\[ \int_0^1 y \, dy = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1}{2} \]
\[ \int_0^1 z \, dz = \left[ z \right]_0^1 = 1 \]

Thus,

\[ E[X] = 12 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot 1 = \frac{12}{8} = \frac{3}{2} = 1.5 \]

### Expected Value of \( Y \)

\[ E[Y] = \iiint_0^1 y \cdot 12x^2yz \, dx \, dy \, dz \]

Separate the integrals:

\[ E[Y] = 12 \int_0^1 \int_0^1 \int_0^1 x^2 y^2 z \, dx \, dy \, dz \]
\[ E[Y] = 12 \left( \int_0^1 x^2 \, dx \right) \left( \int_0^1 y^2 \, dy \right) \left( \int_0^1 z \, dz \right) \]

Compute each integral separately:

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]
\[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \]
\[ \int_0^1 z \, dz = \left[ z \right]_0^1 = 1 \]

Thus,

\[ E[Y] = 12 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 1 = \frac{12}{9} = \frac{4}{3} = 1.33 \]

### Expected Value of \( Z \)

\[ E[Z] = \iiint_0^1 z \cdot 12x^2yz \, dx \, dy \, dz \]

Separate the integrals:

\[ E[Z] = 12 \int_0^1 \int_0^1 \int_0^1 x^2 y z^2 \, dx \, dy \, dz \]
\[ E[Z] = 12 \left( \int_0^1 x^2 \, dx \right) \left( \int_0^1 y \, dy \right) \left( \int_0^1 z^2 \, dz \right) \]

Compute each integral separately:

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]
\[ \int_0^1 y \, dy = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1}{2} \]
\[ \int_0^1 z^2 \, dz = \left[ \frac{z^3}{3} \right]_0^1 = \frac{1}{3} \]

Thus,

\[ E[Z] = 12 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{12}{18} = \frac{2}{3} = 0.67 \]

### Summary

The expected values are:
- \( E[X] = 1.5 \)
- \( E[Y] = 1.33 \)
- \( E[Z] = 0.67 \)

Do you want details or have any questions?

Here are some more questions you might want to explore:
1. What is the variance of each variable \( X \), \( Y \), and \( Z \)?
2. What is the covariance between \( X \) and \( Y \)?
3. What is the probability that \( X + Y + Z \leq 1 \)?
4. How do the marginal density functions of \( X \), \( Y \), and \( Z \) look like?
5. Can you find the moment generating function for \( X \), \( Y \), and \( Z \)?

**Tip:** When calculating expected values, make sure to set up the correct limits of integration and factor the density function correctly for each variable.

Explore how to calculate expected values for variables X, Y, and Z using a joint density function. This problem involves applying the expected value formula to continuous random variables within specified limits. Suitable for advanced mathematics enthusiasts.

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