Question

A biased die is twice as likely to show an even number as an odd number. If such a die is thrown twice, find the probability distribution of the number of sixes. Also, find the mean of the distribution.

Hint:

When working with biased probability distributions, ensure the total probability sums to 1 and carefully calculate probabilities for each outcome.

Answer 1

Step 1: Assign probabilities
Let \( P(3) = P(5) = p \), so \( P(2) = P(4) = P(6) = 2p \).
As the total probability is 1: \[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \implies 9p = 1 \implies p = \frac{1}{9} \] Thus, \( P(6) = 2p = \frac{2}{9} \), and \( P(\text{Not getting six}) = 1 - P(6) = \frac{7}{9} \). 

Step 2: Define the random variable \( X \)
Let \( X \) represent the number of sixes. The possible values of \( X \) are \( 0, 1, 2 \). 

Step 3: Compute probabilities for \( X \)
\[ P(X = 0) = \left( \frac{7}{9} \right)^2 = \frac{49}{81}, \quad P(X = 1) = 2 \cdot \frac{2}{9} \cdot \frac{7}{9} = \frac{28}{81}, \quad P(X = 2) = \left( \frac{2}{9} \right)^2 = \frac{4}{81} \] 

Step 4: Probability distribution of \( X \)
 

\[\begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{49}{81}\\  1 & \frac{28}{81} \\ 2 & \frac{4}{81} \\ \hline \end{array}\]

Step 5: Compute the mean of \( X \)
The mean is given by: \[ \mu = \sum_{i=1}^{3} X_i \cdot P(X_i) = 0 \cdot \frac{49}{81} + 1 \cdot \frac{28}{81} + 2 \cdot \frac{4}{81} = \frac{28}{81} + \frac{8}{81} = \frac{36}{81} = \frac{4}{9} \] 

Step 6: Final result
The probability distribution of \( X \) is: 

\[\begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{49}{81} \\ 1 & \frac{28}{81} \\ 2 & \frac{4}{81} \\ \hline \end{array}\]

The mean of the distribution is \( \frac{4}{9} \).