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Let $X$ be the random variable representing the number written on a randomly selected block. The possible values of $X$ are 0, 1, 2, and 3.
We need to find the probability of each value of $X$. There are a total of 10 blocks.
There are 2 blocks marked with '0'. So, the probability of selecting a block with '0' is: $$P(X=0) = \frac{2}{10} = \frac{1}{5}$$
There are 3 blocks marked with '1'. So, the probability of selecting a block with '1' is: $$P(X=1) = \frac{3}{10}$$
There are 4 blocks marked with '2'. So, the probability of selecting a block with '2' is: $$P(X=2) = \frac{4}{10} = \frac{2}{5}$$
There is 1 block marked with '3'. So, the probability of selecting a block with '3' is: $$P(X=3) = \frac{1}{10}$$
The probability distribution of $X$ is given by: $X = 0, P(X=0) = \frac{1}{5}$ $X = 1, P(X=1) = \frac{3}{10}$ $X = 2, P(X=2) = \frac{2}{5}$ $X = 3, P(X=3) = \frac{1}{10}$
The mean (expected value) of $X$ is given by: $$E(X) = \sum x_i P(X=x_i)$$ $$E(X) = 0 \cdot \frac{1}{5} + 1 \cdot \frac{3}{10} + 2 \cdot \frac{2}{5} + 3 \cdot \frac{1}{10}$$ $$E(X) = 0 + \frac{3}{10} + \frac{4}{5} + \frac{3}{10}$$ $$E(X) = \frac{3}{10} + \frac{8}{10} + \frac{3}{10} = \frac{14}{10} = \frac{7}{5} = 1.4$$
Final Answer: The probability distribution is: $P(X=0) = \frac{1}{5}, P(X=1) = \frac{3}{10}, P(X=2) = \frac{2}{5}, P(X=3) = \frac{1}{10}$. The mean is 1.4.
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