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 probabilities must be 1: \[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \] \[ P(1) + 2p + p + 2p + p + 2p = 1 \] \[ P(1) + 8p = 1 \] Given \( P(1) \) is not explicitly defined, we assume it is \( p \). Then \( p + 8p = 1 \implies 9p = 1 \implies p = \frac{1}{9} \). 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 \( X \) denote the number of sixes observed. The possible values for \( X \) are \( 0, 1, 2 \). 
Step 3: Compute probabilities for \( X \)
The experiment involves two independent trials (implied by the structure of probabilities in Step 5).
\[ P(X = 0) = P(\text{No six in 2 trials}) = \left( \frac{7}{9} \right)^2 = \frac{49}{81} \]\[ P(X = 1) = P(\text{One six in 2 trials}) = \binom{2}{1} P(6) P(\text{Not getting six}) = 2 \cdot \frac{2}{9} \cdot \frac{7}{9} = \frac{28}{81} \]\[ P(X = 2) = P(\text{Two sixes in 2 trials}) = \left( \frac{2}{9} \right)^2 = \frac{4}{81} \] 
Step 4: Probability distribution of \( X \)
The probability distribution of \( X \) is as follows:
\[ \begin{array}{|c|c|} X & P(X) \\ \hline 0 & \frac{49}{81} \\ 1 & \frac{28}{81} \\ 2 & \frac{4}{81} \\ \end{array} \] 
Step 5: Compute the mean of \( X \)
The expected value (mean) of \( X \) is calculated using:
\[ \mu = E(X) = \sum_{i} x_i P(X=x_i) \]\[ \mu = (0 \cdot \frac{49}{81}) + (1 \cdot \frac{28}{81}) + (2 \cdot \frac{4}{81}) \]\[ \mu = 0 + \frac{28}{81} + \frac{8}{81} = \frac{36}{81} \] Simplifying the fraction: \( \mu = \frac{4}{9} \). 
Step 6: Final result
The probability distribution of the random variable \( X \) (number of sixes) is:
\[ \begin{array}{|c|c|} X & P(X) \\ \hline 0 & \frac{49}{81} \\ 1 & \frac{28}{81} \\ 2 & \frac{4}{81} \\ \end{array} \] The mean of this distribution is \( \frac{4}{9} \).