Let $R_1$ be the event that the first ball is red and $R_2$ be the event that the second ball is red. Thus, the probability that both of the balls are red is given by $P(R_1 \cap R_2)$. By the product rule, we have
\[P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1).\]

We have $P(R_1) = 2/5$ because we have a total of five balls of which two are red. Now, if the first ball is red, then the rest are three blue balls and one red ball. Hence the conditional probability $P(R_2 \mid R_1)$ is given by $P(R_2 \mid R_1) = 1/4$.

Combining these, we obtain
\[P(R_1 \cap R_2) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}.\]
In conclusion, the probability that both of the balls are red is $1/10$.

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Definition of Independence
Events […]

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The pass […]