Jack and Jill gamble on a roll of a die, as follows. If the die shows $1$ or $2$ spots, Jack gives Jill $\$1$. If the die shows $5$ or $6$ spots, Jill gives Jack $\$1$. If the die shows $3$ or $4$ spots, no money changes hands. Suppose Jack and Jill play this game $400$ times. The chance that Jill’s net gain is more than $\$20$ is closest to:
a) $10\%$
b) $20\%$
c) $30\%$
d) $40\%$
Addition Rule: $\Pr(A\text{ or }B)= \Pr(A)+\Pr(B)$
$$\Pr(\text{Jack 1 or 2}, + \text{\$1 to Jill})=( 1/6)+(1/6)=2/6$$ $$\Pr(\text{Jill 5 or 6, + \$1 to Jack})=(1/6)+(1/6)=2/6$$ $$\Pr(\text{3 or 4; \$0 both Jack & Jill})= (1/6)+(1/6)=2/6$$
$n=400$
$\Pr(\text{net gain >\$20})$?
In each game, with probability $1/3$ each, Jill wins $\$1$, loses $\$1$ or plays even. Letting $X_i$ denote the gain of Jill in some game $i$, we have
\begin{align} E(X_i) &= \frac{1}{3}(1 + 0 + -1) = 0 \\ \mathrm{Var}(X_i) &= \frac{1}{3}\left(1^2 + 0^2 + (-1)^2\right) = \frac{2}{3} \end{align}
Letting $Y = \sum_{i=1}^{400} X_i$ denote the overall profit of Jill in $n = 400$ games, we can compute the mean and variance of $Y$ as follows, using that the $X_i$ are i.i.d.:
\begin{align} E(Y) &= E\left(\sum_{i=1}^{400} X_i\right) = \sum_{i=1}^{400} E(X_i) = 0 \\ \mathrm{Var}(Y) &= \mathrm{Var}\left(\sum_{i=1}^{400} X_i\right) = \sum_{i=1}^{400} \mathrm{Var}\left(X_i\right) = 400 \cdot \frac{2}{3} = \frac{800}{3}. \end{align}
Using the Gaussian approximation, $Y$ can be approximated by a normally distributed random variable $Y'$ with the same mean and variance, i.e, with $Y'$ satisfying $Y' \sim \mathcal{N}(0, 800/3)$. So we finally get an approximation as
$$P(Y > 20) \approx P(Y' > 20) = P\left(Z > \frac{20}{\sqrt{800/3}}\right) \approx P(Z > 1.22474) \approx 0.11034.$$
(Here $Z$ denotes a standard normal random variable, i.e, $Z \sim \mathcal{N}(0,1)$.)
So the answer is closest to (a) $10\%$.