Solve $P (R_{1} \cap R_{2}), 3 red, 2 blue$

Step-by-step solution

Step 1 · First draw
$P (R_{1}) = \frac{3}{5}$
Three of the five equally likely balls are red.
Step 2 · Update after removal
$P (R_{2} ∣ R_{1}) = \frac{2}{4}$
If the first ball is red, two red balls remain among four. The draws are not independent.
Step 3 · Conditional multiplication
$P (R_{1} \cap R_{2}) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$
Multiply the first probability by the conditional second probability, not by another 3/5.

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More probability

$P (sum = 7)$