Question

Two dice are thrown simultaneously. If $X$ denotes the number of fours, then the expectation of $X$ will be:

• A. $\frac{5}{9}$
• B. $\frac{1}{3}$
• C. $\frac{4}{7}$
• D. $\frac{3}{8}$

Hint:

When calculating the expectation of an event involving multiple independent trials (like rolling dice), the total expectation is simply the sum of the individual expectations. For each die, the expectation is the probability of getting the event (in this case, rolling a four), and since the dice rolls are independent, we can just add the expectations together.

Answer 1

The Correct Option is B

Each die has a probability of \( \frac{1}{6} \) of showing a four. The expectation of \( X \) (the number of fours) is the sum of the expectations for each die:

\[ E(X) = E(X_1) + E(X_2), \]

where:

\[ E(X_1) = \frac{1}{6}, \quad E(X_2) = \frac{1}{6}. \]

Thus:

\[ E(X) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}. \]

Answer 2

Each die has a probability of \( \frac{1}{6} \) of showing a four. The expectation of \( X \) (the number of fours) is the sum of the expectations for each die:

\[ E(X) = E(X_1) + E(X_2) \]

Step 1: Find the expectation for each individual die:

The expectation for each die, \( E(X_1) \) and \( E(X_2) \), is given by the probability of getting a four on each die, which is \( \frac{1}{6} \). Thus, we have:

\[ E(X_1) = \frac{1}{6}, \quad E(X_2) = \frac{1}{6} \]

Step 2: Add the expectations for the two dice:

The total expectation, \( E(X) \), is the sum of the expectations for each die:

\[ E(X) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

Conclusion: Thus, the expectation of \( X \) (the number of fours) is \( \frac{1}{3} \).