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 \( P(3) = P(5) = p \) and \( P(2) = P(4) = P(6) = 2p \). The sum of all probabilities must equal 1: \[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]. Substituting the defined probabilities, we get \( P(1) + 2p + p + 2p + p + 2p = 1 \), which simplifies to \( P(1) + 8p = 1 \). If we assume \( P(1) \) is also related, and given the structure, it's implied \( P(1) = p \). Then \( p + 2p + p + 2p + p + 2p = 1 \implies 9p = 1 \implies p = \frac{1}{9} \). Therefore, \( P(6) = 2p = \frac{2}{9} \), and the probability of not getting a six is \( P(\text{Not getting six}) = 1 - P(6) = 1 - \frac{2}{9} = \frac{7}{9} \).

Step 2: Define the Random Variable \( X \)
Let \( X \) be the random variable representing the number of sixes obtained in two trials. The possible values for \( X \) are \( 0, 1, 2 \). 

Step 3: Calculate Probabilities for \( X \)
The probability of getting zero sixes in two trials is \( P(X = 0) = \left( \frac{7}{9} \right)^2 = \frac{49}{81} \). The probability of getting exactly one six in two trials is \( P(X = 1) = 2 \cdot P(\text{six}) \cdot P(\text{not six}) = 2 \cdot \frac{2}{9} \cdot \frac{7}{9} = \frac{28}{81} \). The probability of getting two sixes in two trials is \( P(X = 2) = \left( \frac{2}{9} \right)^2 = \frac{4}{81} \). 

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 expected value (mean) of \( X \) is calculated as: \[ \mu = \sum x \cdot P(x) = (0 \cdot \frac{49}{81}) + (1 \cdot \frac{28}{81}) + (2 \cdot \frac{4}{81}) = 0 + \frac{28}{81} + \frac{8}{81} = \frac{36}{81} = \frac{4}{9} \] 

Step 6: Final Results
The probability distribution of the random variable \( X \) (number of sixes in two trials) 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 this distribution is \( \frac{4}{9} \).