A bag has 7 red and 5 black balls. Two balls are drawn one after the other without replacement. What is the probability that both are red?

Explanation

1. 01

   Let $R_1$, $R_2$ be red on the first and second draws: $P(R_1) = \frac{7}{12}$, and with 6 red of 11 left, $P(R_2 \mid R_1) = \frac{6}{11}$.

2. 02

   $\frac{7}{12} \times \frac{6}{11} = \frac{7}{22}$.