爱明和舒婷在玩一个电脑游戏，她们各自向目标射击。对于每次射击，两人击中目标的概率均为 0.25，且独立于两人的所有其他射击，并且她们对目标的尝试次数相同。求：

1. Define the random variable and its distribution.

Let $X$ be the number of shots, out of 20, that hit the target.
$X \sim \mathrm{B}(20, 0.25)$

2. Compare the probabilities of hitting the target for values around the expected mean to find the most likely outcome.

The most likely number corresponds to the maximum probability. We can check the probabilities for values around the mean $\mathrm{E}(X) = 20 \times 0.25 = 5$ using a calculator:
$\mathrm{P}(X = 4) \approx 0.1897$
$\mathrm{P}(X = 5) \approx 0.2023$
$\mathrm{P}(X = 6) \approx 0.1686$
Since $\mathrm{P}(X = 5)$ is the highest, the most likely number of shots is 5.

1. Define the random variable and set up the expression for conditional probability.

Let $X$ be the number of shots, out of $20$, that Ai Ming hits the target.
$X \sim \mathrm{B}(20, 0.25)$

We need to find $\mathrm{P}(X \ge 5 \mid X \le 10)$.
$\mathrm{P}(X \ge 5 \mid X \le 10) = \frac{\mathrm{P}(X \ge 5 \text{ and } X \le 10)}{\mathrm{P}(X \le 10)}$

2. Express the numerator as an interval and convert it into a difference of cumulative probabilities.

$= \frac{\mathrm{P}(5 \le X \le 10)}{\mathrm{P}(X \le 10)}$
$= \frac{\mathrm{P}(X \le 10) - \mathrm{P}(X \le 4)}{\mathrm{P}(X \le 10)}$

3. Evaluate the probabilities using a graphing calculator and compute the final answer.

$= \frac{0.986139 - 0.414842}{0.986139}$
$= \frac{0.571297}{0.986139}$
$= 0.579327 \approx 0.579 \text{ (3 s.f.)}$

1. Find the probability that a player takes exactly $k$ shots to hit the target for the first time.

Let $X$ and $Y$ be the number of shots taken by Ai Ming and Shu Ting to hit the target respectively.
$P(X=k) = (1 - 0.25)^{k-1}(0.25) = (0.75)^{k-1}(0.25)$

2. Find the probability that both players take exactly $k$ shots to hit the target.

Since their shots are independent, the probability that both take exactly $k$ shots is:
$P(X=k) \times P(Y=k) = \left( (0.75)^{k-1}(0.25) \right)^2$
$= (0.25)^2 (0.75^2)^{k-1} = \frac{1}{16} \left( \frac{9}{16} \right)^{k-1}$

3. Sum the probabilities over all possible values of $k$ to infinity.

The total probability is the sum of these probabilities for $k=1, 2, 3, \dots$
$P(X=Y) = \sum_{k=1}^{\infty} \frac{1}{16} \left( \frac{9}{16} \right)^{k-1}$
This is an infinite geometric series with first term $a = \frac{1}{16}$ and common ratio $r = \frac{9}{16}$.
$P(X=Y) = \frac{a}{1 - r} = \frac{\frac{1}{16}}{1 - \frac{9}{16}} = \frac{\frac{1}{16}}{\frac{7}{16}} = \frac{1}{7}$