Mathematics 9709/72 — February/March 2016

Walkthrough

First identify what each variable represents. $X$ is the number of sixes in 20 throws, so each throw has probability $p=\frac{1}{6}$ of success. $Y$ is the number of odd-numbered scores; on a die, 1, 3 and 5 are odd, so $p=\frac{1}{2}$. Both are binomial counts because each throw is independent and has a fixed number of trials.

For a binomial distribution, the mean is $np$ and the variance is $np(1-p)$. Apply this to $X$ and $Y$ separately. Then, because the two dice are thrown independently, $X$ and $Y$ are independent random variables. For independent variables, expectations add and variances add. This is why we add the two variances before taking the square root to find the standard deviation.

Key Takeaways

• A count of successes in a fixed number of independent trials is binomial.
• For $X \sim \mathrm{Bin}(n,p)$, $E(X)=np$ and $\mathrm{Var}(X)=np(1-p)$.
• For independent random variables, $E(X+Y)=E(X)+E(Y)$ and $\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)$.
• Standard deviation is the positive square root of variance.

Common Mistakes

• Using $p=\frac{1}{6}$ for $Y$ as well; odd-numbered scores have probability $\frac{1}{2}$.
• Adding standard deviations instead of variances. You must add variances first, then square root.
• Forgetting that the variance addition rule requires independence. Here the dice are separate, so independence is valid.
• Giving the variance instead of the standard deviation as the final answer.

Things to Be Careful About

• Keep exact fractions throughout and only round at the end.
• Check the probability for $Y$: odd scores are 1, 3, 5, so $p=\frac{3}{6}=\frac{1}{2}$.
• The final standard deviation can be written as $\frac{\sqrt{70}}{3}$, $\sqrt{\frac{70}{9}}$, or $2.79$ to 3 significant figures.
• The mark scheme accepts equivalent forms such as $\sqrt{2}\times\frac{5}{3}$ for the standard deviation.