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Let $P(\text{odd}) = p$. Then $P(\text{even}) = 2p$. Since the sum of probabilities must be 1, we have: $P(\text{odd}) + P(\text{even}) = 1$ $p + 2p = 1$ $3p = 1$ $p = \frac{1}{3}$ So, $P(\text{odd}) = \frac{1}{3}$ and $P(\text{even}) = \frac{2}{3}$.
Since 6 is an even number, the probability of rolling a 6 is $\frac{2}{3}$ divided by the number of even numbers (2, 4, 6), which is 3. Thus, $P(6) = \frac{P(\text{even})}{3} = \frac{2/3}{3} = \frac{2}{9}$ The probability of not rolling a 6 is $P(\text{not 6}) = 1 - P(6) = 1 - \frac{2}{9} = \frac{7}{9}$.
Let $X$ be the number of sixes in two throws. $X$ can take values 0, 1, or 2.
We use the binomial probability formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $n=2$ and $p = \frac{2}{9}$. $P(X=0) = \binom{2}{0} (\frac{2}{9})^0 (\frac{7}{9})^2 = 1 \cdot 1 \cdot \frac{49}{81} = \frac{49}{81}$ $P(X=1) = \binom{2}{1} (\frac{2}{9})^1 (\frac{7}{9})^1 = 2 \cdot \frac{2}{9} \cdot \frac{7}{9} = \frac{28}{81}$ $P(X=2) = \binom{2}{2} (\frac{2}{9})^2 (\frac{7}{9})^0 = 1 \cdot \frac{4}{81} \cdot 1 = \frac{4}{81}$
The probability distribution is: $X = 0: P(X=0) = \frac{49}{81}$ $X = 1: P(X=1) = \frac{28}{81}$ $X = 2: P(X=2) = \frac{4}{81}$
The mean $\mu$ is given by $\mu = \sum x_i P(X=x_i)$. $\mu = 0 \cdot \frac{49}{81} + 1 \cdot \frac{28}{81} + 2 \cdot \frac{4}{81} = 0 + \frac{28}{81} + \frac{8}{81} = \frac{36}{81} = \frac{4}{9}$
Final Answer: Probability Distribution: P(X=0) = 49/81, P(X=1) = 28/81, P(X=2) = 4/81. Mean = 4/9
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