一轮游戏是通过沿着如下图所示的网格线，将棋子从起点 $X$ 随机移动到终点 $A$ 至 $E$ 中的一个来进行的。

• 棋子只能沿着网格线向上或向右移动。
• 在离开 $X$ 后的每一步移动中，玩家随机将棋子向右移动的概率为 $p$，向上移动的概率为 $q$，且 $p+q=1$。在每个分叉点所做出的移动是相互独立的。
• 玩家要赢得一轮游戏，棋子必须经过 $Y$ 并最终到达点 $B$。

1. Find the probability of reaching Y from X

From $X(0,0)$ to $Y(4,2)$ the counter must make $4$ moves right and $2$ moves up, in any order. The number of such routes is $\binom{6}{2}=15$. Each route consists of $4$ right moves (each probability $p$) and $2$ up moves (each probability $q$), so each route has probability $p^4q^2$.

$P(X\to Y) = 15\,p^4q^2$

2. Find the probability of going from Y to B

From $Y(4,2)$ the counter must finish on the right boundary at $B(6,3)$. To land on $B$ it must reach the junction $(5,3)$ — one step to the left of $B$ — and then take the final rightward step onto the boundary, which is the forced exit move (probability $1$, no factor). From $Y(4,2)$ to $(5,3)$ requires $1$ right and $1$ up move in either order: $\binom{2}{1}=2$ routes, each of probability $pq$.

$P(Y\to B) = 2\,pq$

3. Combine the two independent stages

Since the two stages are independent, multiply:

$P(\text{win}) = P(X\to Y)\times P(Y\to B) = 15p^4q^2 \times 2pq = 30p^5q^3.$ (shown)

1. Calculate the probability of winning a single round.

Given $p = \frac{4}{5}$, we have $q = 1 - p = \frac{1}{5}$.
Using the result shown in part (a), the probability of winning a single round is:
$P(\text{win}) = 30p^5q^3 = 30 \left(\frac{4}{5}\right)^5 \left(\frac{1}{5}\right)^3 = \frac{6144}{78125} = 0.0786432$

2. Set up the binomial distribution for the number of rounds won.

Let $W$ be the number of rounds won out of 15.
Since each round is independent and has two possible outcomes (win or lose) with a constant probability of success, $W$ follows a binomial distribution:
$W \sim \mathrm{B}(15, 0.0786432)$

3. Calculate the probability of winning at least one-third of the rounds.

One-third of 15 rounds is $5$ rounds. We need to find the probability that Wang wins at least 5 rounds, which is $P(W \ge 5)$.
$P(W \ge 5) = 1 - P(W \le 4)$
Using a graphing calculator:
$P(W \ge 5) \approx 1 - 0.996419 = 0.0035811 \approx 0.00358 \text{ (to 3 s.f.)}$

1. Find the mean and variance of the number of rounds won in one game

For one round, the probability of winning is
$p_{\text{win}} = 30p^5q^3 = 30\left(\tfrac{4}{5}\right)^5\left(\tfrac{1}{5}\right)^3 = \frac{30720}{390625} = 0.0786432.$
In one game ($15$ independent rounds), the number of rounds won is $W \sim \mathrm{B}(15,\,0.0786432)$, so
$\mathrm{E}(W) = 15(0.0786432) = 1.17965,$
$\mathrm{Var}(W) = 15(0.0786432)(0.9213568) = 1.08688.$

2. Apply the Central Limit Theorem to the sample mean over 40 games

Let $\bar{W}$ be the mean number of rounds won per game over the $40$ games. Since $n = 40 \ge 30$, by the Central Limit Theorem $\bar{W}$ is approximately normal:
$\bar{W} \sim \mathrm{N}\!\left(\mathrm{E}(W),\,\frac{\mathrm{Var}(W)}{40}\right) = \mathrm{N}\!\left(1.17965,\,\frac{1.08688}{40}\right) = \mathrm{N}(1.17965,\,0.0271719)$
approximately.

3. Evaluate the required probability

$P(\bar{W} \le 1) = P\!\left(Z \le \frac{1 - 1.17965}{\sqrt{0.0271719}}\right) = P(Z \le -1.0898) \approx 0.138.$
(Using the GC: $\mathrm{normalcdf}(-10^{99},\,1,\,1.17965,\,0.164839) = 0.1379$.)