Serene plays a simple online game which moves a counter one step at a time along a straight line. The counter starts from point S. In each turn, the counter moves one step forward with probability $\frac{3}{5}$ or one step backward with probability $\frac{2}{5}$. The online game ends after 10 turns. Let $X$ be the number of turns, out of 10, where the counter moves one step forward.

1. Explain why $X = 8$ results in a net of 6 steps forward.

The total number of turns is 10. If the counter moves forward $X = 8$ times, it must move backward for the remaining $10 - 8 = 2$ turns.
The net movement from the starting point S is $8 - 2 = 6$ steps forward. Hence, the counter ends up exactly 6 steps ahead of S.

2. Formulate the probability expression using the binomial distribution.

The number of forward steps $X$ follows a binomial distribution $X \sim \text{B}(10, \frac{3}{5})$.
The probability that the counter ends up 6 steps ahead is $P(X = 8)$.
$P(X = 8) = \binom{10}{8} \left(\frac{3}{5}\right)^8 \left(\frac{2}{5}\right)^2$

3. Evaluate the probability to 3 significant figures.

$P(X = 8) = 45 \times (0.6)^8 \times (0.4)^2$
$P(X = 8) = 0.120932...$
$P(X = 8) \approx 0.121 \text{ (to 3 s.f.)}$

1. Determine the required number of forward steps ($X$) to win the game.

Let $D$ be the number of steps the counter ends up ahead of point S. Since the counter moves backward for the remaining $10 - X$ turns,
$D = X - (10 - X) = 2X - 10$
For the player to win, the counter must end up at least 6 steps ahead, so $D \ge 6$:
$2X - 10 \ge 6 \implies 2X \ge 16 \implies X \ge 8$

2. Model $X$ with a binomial distribution.

Since there are 10 independent turns and the probability of moving forward in each turn is $\frac{3}{5} = 0.6$, $X$ follows a binomial distribution:
$X \sim \mathrm{B}(10, 0.6)$
The probability of winning is $\mathrm{P}(X \ge 8)$.

3. Calculate $\mathrm{P}(X \ge 8)$ to show the final result.

$\begin{aligned} \mathrm{P}(X \ge 8) &= \mathrm{P}(X = 8) + \mathrm{P}(X = 9) + \mathrm{P}(X = 10) \\ &= \binom{10}{8}(0.6)^8(0.4)^2 + \binom{10}{9}(0.6)^9(0.4)^1 + \binom{10}{10}(0.6)^{10}(0.4)^0 \\ &= 0.120932 + 0.040311 + 0.006047 \\ &= 0.167290 \\ &\approx 0.167 \text{ (to 3 s.f.)} \end{aligned}$

1. Determine the required number of wins.

Number of games won $= 15\% \text{ of } 20 = 0.15 \times 20 = 3$

2. Define the binomial random variable and compute the probability using the unrounded probability from part (b).

Let $Y$ be the number of online games won out of $20$.
$Y \sim \mathrm{B}(20, 0.16729)$

1. Calculate the total probability that Serene wins a game, incorporating the unrounded probability of winning the online game from part (b).

Let $O$ be the event she plays the online game and $W$ be the event she plays the word puzzle game.
Let $\text{Win}$ be the event she wins the game.
Using the unrounded probability from part (b), $P(\text{Win} \mid O) \approx 0.16729$.

$P(\text{Win}) = P(W)P(\text{Win} \mid W) + P(O)P(\text{Win} \mid O)$
$P(\text{Win}) = (0.7 \times 0.4) + (0.3 \times 0.16729)$
$P(\text{Win}) = 0.28 + 0.050187 = 0.330187$

2. Use the conditional probability formula to find the probability she played the word puzzle game given that she won.

$P(W \mid \text{Win}) = \frac{P(W \cap \text{Win})}{P(\text{Win})}$
$P(W \mid \text{Win}) = \frac{0.28}{0.330187}$
$P(W \mid \text{Win}) = 0.84797 \dots \approx 0.848 \text{ (to 3 s.f.)}$