2022 H2 Maths Paper 2 Q6 — Worked Solution

Part (a)

3 marks

Approach

Anil’s first throw only sets a target number; it cannot itself win. From then on the two players alternate attempts to match the number thrown immediately before them: Babs’s 1st attempt, then Anil’s 1st attempt, then Babs’s 2nd, and so on. Each attempt is independent, succeeding with probability $\frac{1}{6}$ and failing with probability $\frac{5}{6}$ . Anil wins exactly when the first success in this alternating sequence falls on one of his attempts.

Anil wins on his $n$ th attempt ( $n = 1,2,3,\ldots$ ) if Babs’s 1st through $n$ th attempts all fail, Anil’s 1st through $(n - 1)$ th attempts all fail, and Anil’s $n$ th attempt succeeds — a total of $2n - 1$ failures followed by one success:

P\left( \text{Anil wins on his }n\text{th attempt} \right) = \left( \frac{5}{6} \right)^{2n - 1}\frac{1}{6}

Summing over all $n$ (a geometric series with first term $\frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ and common ratio $\left( \frac{5}{6} \right)^{2} = \frac{25}{36}$ ),

P\left( \text{Anil wins} \right) = \sum_{n = 1}^{\infty}\left( \frac{5}{6} \right)^{2n - 1}\frac{1}{6}

= \frac{5/36}{1 - 25/36}

= \frac{5/36}{11/36}

= \frac{5}{11}

To check your answer: by the same method,

P\left( \text{Babs wins} \right) = \sum_{m = 1}^{\infty}\left( \frac{5}{6} \right)^{2(m - 1)}\frac{1}{6} = \frac{1/6}{1 - 25/36} = \frac{6}{11}

, and

\frac{5}{11} + \frac{6}{11} = 1

, confirming the game ends with probability

1

and the two results are consistent ✓

$\therefore\ P\left( \text{Anil wins} \right) = \frac{5}{11}$

Part (b)

3 marks

Approach

We need $P\left( \text{Babs wins on her 2nd throw} \mid \text{Babs wins} \right)$ , so we need both the probability that Babs wins specifically on her 2nd attempt, and $P\left( \text{Babs wins} \right)$ from (a).

Babs wins on her $m$ th attempt ( $m = 1,2,3,\ldots$ ) if Babs’s 1st through $(m - 1)$ th attempts all fail, Anil’s 1st through $(m - 1)$ th attempts all fail, and Babs’s $m$ th attempt succeeds — a total of $2(m - 1)$ failures followed by one success:

P\left( \text{Babs wins on her }m\text{th attempt} \right) = \left( \frac{5}{6} \right)^{2(m - 1)}\frac{1}{6}

For $m = 2$ (Babs’s second throw is her second attempt),

P\left( \text{Babs wins on her 2nd attempt} \right) = \left( \frac{5}{6} \right)^{2} \times \frac{1}{6}

= \frac{25}{216}

From (a), $P\left( \text{Babs wins} \right) = \frac{6}{11}$ . By conditional probability,

P\left( \text{Babs wins on 2nd throw} \mid \text{Babs wins} \right) = \frac{P\left( \text{Babs wins on 2nd attempt} \right)}{P\left( \text{Babs wins} \right)}

= \frac{25/216}{6/11}

= \frac{25}{216} \times \frac{11}{6}

= \frac{275}{1296}

To check your answer:

\frac{275}{1296} \approx 0.212

; since Babs’s chance of winning on a given attempt decreases geometrically, this should be a little less than

P\left( \text{Babs wins on 1st attempt} \mid \text{Babs wins} \right) = \frac{1/6}{6/11} = \frac{11}{36} \approx 0.306

— and indeed it is ✓

$\therefore\ P\left( \text{Babs wins on her 2nd throw} \mid \text{Babs wins} \right) = \frac{275}{1296}$

2022 H2 Maths Paper 2, Question 6

Part (a)

Part (b)

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