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The sum of all probabilities in a probability distribution must equal 1. Therefore, we have: $p + 2p + 3p + p = 1$
Combining the terms, we get: $7p = 1$ Dividing both sides by 7, we find: $p = \frac{1}{7}$
The mean (or expected value) of a discrete random variable X is given by: $E(X) = \sum [X \cdot P(X)]$
Substituting the given values and the calculated value of $p$, we have: $E(X) = (0 \cdot p) + (2 \cdot 2p) + (4 \cdot 3p) + (5 \cdot p)$ $E(X) = 0 + 4p + 12p + 5p$ $E(X) = 21p$
Since $p = \frac{1}{7}$, we substitute this value into the expression for $E(X)$: $E(X) = 21 \cdot \frac{1}{7}$ $E(X) = 3$
Final Answer: p = 1/7, Mean = 3
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