2022 H2 Mathematics Paper 2 Question 6

Probability

Answers

(a)

511.{\frac{5}{11}.}

(b)

2751296.{\frac{275}{1296}.}

Full solutions

(a)

Probability Anil wins=(56)(16)+(56)3(16)+(56)5(16)+…=(56)(16)1−(56)2=511  ■\begin{align*} & \textrm{Probability Anil wins} \\ &= \left( \frac{5}{6} \right) \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^3 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^5 \left( \frac{1}{6} \right) + \ldots \\ &= \frac{\left( \frac{5}{6} \right)\left( \frac{1}{6} \right)}{1- \left( \frac{5}{6} \right)^2} \\ &= \frac{5}{11} \; \blacksquare \end{align*}

(b)

P(Wins on second∣Babs win)=P(Wins on second∩Babs win)P(Babs win)=(56)2(16)1−511=2751296  ■\begin{align*} & \mathrm{P}\left( \textrm{Wins on second} \mid \textrm{Babs win} \right) \\ &= \frac{\mathrm{P}\left( \textrm{Wins on second} \cap \textrm{Babs win} \right)}{\mathrm{P}\left(\textrm{Babs win} \right)} \\ &= \frac{\left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)}{1-\frac{5}{11}} \\ &= \frac{275}{1296} \; \blacksquare \end{align*}

Question Commentary

For part (a), we should first identify the first few cases in which Anil can win. He can win on his first throw, or his second throw (provided Babs does not win), etc, and we should write out the probability for each case. We then observe that this game can potentially go on forever and apply our sum to infinity formula from the topic of geometric progressions to get the answer.

For part (b), we need to spot the crucial key word "given" and apply our conditional probability formula to get the answer.