A bag contains 5 red marbles and 5 blue marbles. A marble is drawn, its color noted, and then replaced before a second marble is drawn.…

Because the marble is replaced, the two draws are independent and the second draw has the same chances as the first. The probability of red on each draw is $\frac{5}{10}=\frac{1}{2}$, so the probability of two reds is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$, choice A. Choice B ($\frac{1}{2}$) is the probability of a single red draw. Choice C ($\frac{2}{5}$) and choice D ($\frac{9}{20}$) do not come from the replacement setup. Without replacement, the answer would be $\frac{5}{10}\cdot\frac{4}{9}=\frac{2}{9}$, but replacement keeps both draws at $\frac{1}{2}$.