A bag contains 4 red marbles and 6 blue marbles. Two marbles are drawn without replacement. What is the probability that the first is red…

The draws are without replacement, so the second probability depends on the first. The probability that the first marble is red is $\frac{4}{10}$. Given that, $9$ marbles remain and $6$ of them are blue, so the probability the second is blue is $\frac{6}{9}$. Multiply: $\frac{4}{10}\cdot\frac{6}{9}=\frac{24}{90}=\frac{4}{15}$, choice F. Choice H ($\frac{6}{25}$) treats the draws as independent, $\frac{4}{10}\cdot\frac{6}{10}$, as if the marble were replaced. Choice G ($\frac{2}{5}$) is only the probability that the first is red. Choice J ($\frac{8}{15}$) is double the correct value.