Ai Ming plays a computer game with Shu Ting, where they each shoot at a target. For each shot, the probability that the shot hits the target is 0.25, independently of all other shots for both players and they each take the same number of attempts at the target. Find

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 $r = \frac{9}{16}$.
$P(X=Y) = \frac{a}{1 - r} = \frac{\frac{1}{16}}{1 - \frac{9}{16}} = \frac{\frac{1}{16}}{\frac{7}{16}} = \frac{1}{7}$