Let's solve Question 4 (a-e) based on the provided PMF table.

Given Joint PMF Table for $X$, $Y$, $Z$:

• We use the joint probabilities from the table to calculate the required probabilities.

(a) $P(Y = 2)$:

To calculate $P(Y = 2)$, we sum all probabilities where $Y = 2$: $P(Y = 2) = P(X=1, Y=2, Z=1) + P(X=2, Y=2, Z=1) + P(X=3, Y=2, Z=1) + \cdots$

From the table: $P(Y = 2) = (0.006 + 0.005 + 0.004) + (0.02 + 0.145 + 0.04) + (0.02 + 0.08 + 0.05)$ $P(Y = 2) = 0.015 + 0.205 + 0.15 = 0.37$

(b) $P(Z = 2)$:

To calculate $P(Z = 2)$, sum all probabilities where $Z = 2$: $P(Z = 2) = P(X=1, Y=1, Z=2) + P(X=2, Y=1, Z=2) + \cdots$

From the table: $P(Z = 2) = (0.07 + 0.06 + 0.01) + (0.02 + 0.08 + 0.05) + (0.09 + 0.04 + 0.03)$ $P(Z = 2) = 0.14 + 0.15 + 0.16 = 0.45$

(c) $P(X = 2)$:

To calculate $P(X = 2)$, sum all probabilities where $X = 2$: $P(X = 2) = P(X=2, Y=1, Z=1) + P(X=2, Y=2, Z=1) + P(X=2, Y=3, Z=1) + \cdots$

From the table: $P(X = 2) = (0.002 + 0.005 + 0.008) + (0.06 + 0.08 + 0.04) + (0.01 + 0.145 + 0.06)$ $P(X = 2) = 0.015 + 0.18 + 0.215 = 0.41$

(d) $P(Y = X)$:

Here, $Y = X$ means we only consider diagonal elements $(X=1, Y=1), (X=2, Y=2), (X=3, Y=3)$ for all $Z$.

From the table: $P(Y = X) = P(X=1, Y=1, Z=1) + P(X=2, Y=2, Z=1) + P(X=3, Y=3, Z=1)$ $+ P(X=1, Y=1, Z=2) + P(X=2, Y=2, Z=2) + P(X=3, Y=3, Z=2)$ $+ P(X=1, Y=1, Z=3) + P(X=2, Y=2, Z=3) + P(X=3, Y=3, Z=3)$

$P(Y = X) = (0.001 + 0.005 + 0.009) + (0.07 + 0.08 + 0.03) + (0.03 + 0.145 + 0.08)$ $P(Y = X) = 0.015 + 0.18 + 0.255 = 0.45$

(e) $P(X + Y = 4)$:

Here, $X + Y = 4$ means we consider:

• $(X=1, Y=3)$
• $(X=2, Y=2)$
• $(X=3, Y=1)$ for all $Z$.

From the table: $P(X + Y = 4) = P(X=1, Y=3, Z=1) + P(X=1, Y=3, Z=2) + P(X=1, Y=3, Z=3)$ $+ P(X=2, Y=2, Z=1) + P(X=2, Y=2, Z=2) + P(X=2, Y=2, Z=3)$ $+ P(X=3, Y=1, Z=1) + P(X=3, Y=1, Z=2) + P(X=3, Y=1, Z=3)$

$P(X + Y = 4) = (0.007 + 0.09 + 0.07) + (0.005 + 0.08 + 0.145) + (0.003 + 0.01 + 0.05)$ $P(X + Y = 4) = 0.167 + 0.23 + 0.063 = 0.46$

Final Answers:

(a) $P(Y = 2) = 0.37$
(b) $P(Z = 2) = 0.45$
(c) $P(X = 2) = 0.41$
(d) $P(Y = X) = 0.45$
(e) $P(X + Y = 4) = 0.46$

Would you like me to explain any of these calculations in further detail?

Here are five related questions you might explore:

1. How do you compute the marginal PMFs $P(X)$, $P(Y)$, or $P(Z)$?
2. How can conditional probabilities, such as $P(X | Y=2)$, be calculated?
3. How do the joint probabilities relate to the independence of $X$, $Y$, and $Z$?
4. What is the expected value of $X$, $Y$, or $Z$?
5. How do you calculate $P(X + Y > 4)$?

Tip: Always verify that the sum of all joint probabilities equals 1, as this ensures the PMF is valid.