Re-establishment of Hardy Weinberg Equilibrium After Selection
Revised to add complete solution

Before the owls arrive, 100 out of the 2500 mice are dark-colored. This ratio provides the frequency of the recessive allele

$$q^2 = \frac{100}{2500}$$

So $q$ is 10/50 or 0.2. This means that the frequency of the dominant allele $p$ is 1 - 0.2 or 0.8.

Knowing these frequencies, it is possible to calculate the number of each allele

$$AA = (0.8)^2 \times 2500 = 1600$$
$$Aa = 2 (0.8) (0.2) \times 2500 = 800$$
$$aa = (0.2)^2 \times 2500 = 100$$

Of the light-colored mice, 2/3 are AA and 1/3 are Aa. The same proportions hold after the owls leave, but the total humber has been reduced. The dark-colored mice are all aa.

$$AA = \frac{2}{3} \times 240 = 160$$
$$Aa = \frac{1}{3} \times 240 = 80$$
$$aa = 80$$

From these values, the total number of each allele may be calculated

$$A = 2 \times 160 + 80 = 400$$
$$a = 80 + 2 \times 80 = 240$$

These numbers provide the frequency of each allele post-selection

$$p = \frac{400}{640} \approx 0.63$$
$$q = \frac{240}{640} \approx 0.37$$

Applying Hardy Weinberg for the new equilibrium

$$aa \approx (0.37)^2 \approx 0.14 $$

which is, "approximately" 15%.