We are tasked with finding the expected value of $Z$, given that $Y = 10$, based on the joint probability distribution.

Let's analyze the provided table first. We know that:

• $Y$ takes values 10 and 20.
• $Z$ takes values 29 and 21.

The relevant row for $Y = 10$ has the probabilities:

• $P(Y = 10, Z = 29) = 0.2$
• $P(Y = 10, Z = 21) = 0.1$

Step 1: Normalize the conditional probabilities

We first need to compute $P(Z = z \mid Y = 10)$, which is the probability of $Z$ given that $Y = 10$.

$P(Z = 29 \mid Y = 10) = \frac{P(Y = 10, Z = 29)}{P(Y = 10)} = \frac{0.2}{0.2 + 0.1} = \frac{0.2}{0.3} = \frac{2}{3}$ $P(Z = 21 \mid Y = 10) = \frac{P(Y = 10, Z = 21)}{P(Y = 10)} = \frac{0.1}{0.3} = \frac{1}{3}$

Step 2: Compute the expected value

The expected value of $Z$, given that $Y = 10$, is calculated as:

$E[Z \mid Y = 10] = P(Z = 29 \mid Y = 10) \times 29 + P(Z = 21 \mid Y = 10) \times 21$ Substituting the values:

$E[Z \mid Y = 10] = \frac{2}{3} \times 29 + \frac{1}{3} \times 21$

Now, let's calculate this.The expected value of $Z$, given that $Y = 10$, is approximately 26.33.

Would you like to delve deeper into any of the steps? Here are five related questions that can expand your understanding:

1. How would you calculate the marginal probability $P(Z = 29)$?
2. What is the expected value of $Z$ given $Y = 20$?
3. How do you calculate conditional probability in general?
4. Can the expected value change if the joint probabilities for $Z$ are different?
5. What is the significance of the joint probability distribution in statistics?

Tip: Always ensure to normalize probabilities when working with conditional expectations to get accurate results.