A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random without replacement. What is the probability that both…

The marbles are drawn without replacement, so the second probability depends on the first draw. The chance the first marble is red is $\frac{5}{8}$. If it was red, 4 red marbles remain among 7, so the chance the second is also red is $\frac{4}{7}$. Multiply: $\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$. A common wrong answer is $\frac{25}{64}$, which uses $\frac{5}{8}$ twice, as if the first marble were replaced. Another is $\frac{5}{7}$, which uses only the second draw's counts. Always update the counts after each draw.