CAIE S1 2022 November — Question 7 8 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2022
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeSampling without replacement from bags/boxes
DifficultyStandard +0.8 This is a multi-part conditional probability question requiring tree diagrams and careful tracking of game states across multiple turns. Part (c) involves reverse conditional probability (given Tom wins on turn 2, what's P(Sam's first was red)?), which requires Bayes' theorem thinking and is significantly harder than standard sampling problems. The sequential nature with winning conditions adds complexity beyond routine 'draw without replacement' exercises.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles

7 Sam and Tom are playing a game which involves a bag containing 5 white discs and 3 red discs. They take turns to remove one disc from the bag at random. Discs that are removed are not replaced into the bag. The game ends as soon as one player has removed two red discs from the bag. That player wins the game. Sam removes the first disc.
  1. Find the probability that Tom removes a red disc on his first turn.
  2. Find the probability that Tom wins the game on his second turn.
  3. Find the probability that Sam removes a red disc on his first turn given that Tom wins the game on his second turn.
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Question 7(a):
AnswerMarks Guidance
\([P(\text{SR TR}) + P(\text{SW TR})] = \frac{3}{8}\times\frac{2}{7} + \frac{5}{8}\times\frac{3}{7}\)M1 \(\frac{3}{8}\times\frac{2}{7} + k\) or \(l + \frac{5}{8}\times\frac{3}{7}\), \(0 < k, l < 1\)
\(= \frac{21}{56}, \frac{3}{8},\ 0.375\)A1 SC B1 for \(\frac{3}{8}\) with no explanation
Question 7(b):
\([\text{RRWR, WRRR, WRWR}]\)
AnswerMarks Guidance
\(\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{1}{5} + \frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5} + \frac{5}{8}\times\frac{3}{7}\times\frac{4}{6}\times\frac{2}{5}\)M1 \(\frac{m}{8}\times\frac{n}{7}\times\frac{o}{6}\times\frac{q}{5}\), \(1 \leqslant m,n,o,q \leqslant 5\), \(m\neq n\neq o\neq q\)
\(\left[= \frac{1}{56} + \frac{1}{56} + \frac{1}{14}\right]\)A1 Probability for one scenario correct, accept unsimplified
M1Adding probabilities for 3 correct scenarios and no incorrect
\(= \frac{180}{1680}, \frac{3}{28},\ 0.107\)A1 SC B1 for \(\frac{3}{28}\) with inadequate explanation
Question 7(c):
AnswerMarks Guidance
\([P(\text{S first disc R} \mid T2)] = \frac{\dfrac{30}{1680}}{\dfrac{3}{28}} = \frac{1}{56} \div \frac{3}{28}\)M1 their \(P(\text{RRWR})\) or \(\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{1}{5}\) divided by their 7(b) — must be a prob or \(\frac{3}{28}\)
\(\frac{1}{6},\ 0.167\)A1
# Question 7(a):

$[P(\text{SR TR}) + P(\text{SW TR})] = \frac{3}{8}\times\frac{2}{7} + \frac{5}{8}\times\frac{3}{7}$ | M1 | $\frac{3}{8}\times\frac{2}{7} + k$ or $l + \frac{5}{8}\times\frac{3}{7}$, $0 < k, l < 1$

$= \frac{21}{56}, \frac{3}{8},\ 0.375$ | A1 | **SC B1** for $\frac{3}{8}$ with no explanation

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# Question 7(b):

$[\text{RRWR, WRRR, WRWR}]$

$\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{1}{5} + \frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5} + \frac{5}{8}\times\frac{3}{7}\times\frac{4}{6}\times\frac{2}{5}$ | M1 | $\frac{m}{8}\times\frac{n}{7}\times\frac{o}{6}\times\frac{q}{5}$, $1 \leqslant m,n,o,q \leqslant 5$, $m\neq n\neq o\neq q$

$\left[= \frac{1}{56} + \frac{1}{56} + \frac{1}{14}\right]$ | A1 | Probability for one scenario correct, accept unsimplified

| M1 | Adding probabilities for 3 correct scenarios and no incorrect

$= \frac{180}{1680}, \frac{3}{28},\ 0.107$ | A1 | **SC B1** for $\frac{3}{28}$ with inadequate explanation

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# Question 7(c):

$[P(\text{S first disc R} \mid T2)] = \frac{\dfrac{30}{1680}}{\dfrac{3}{28}} = \frac{1}{56} \div \frac{3}{28}$ | M1 | their $P(\text{RRWR})$ or $\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{1}{5}$ divided by their 7(b) — must be a prob or $\frac{3}{28}$

$\frac{1}{6},\ 0.167$ | A1 |
7 Sam and Tom are playing a game which involves a bag containing 5 white discs and 3 red discs. They take turns to remove one disc from the bag at random. Discs that are removed are not replaced into the bag. The game ends as soon as one player has removed two red discs from the bag. That player wins the game.

Sam removes the first disc.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Tom removes a red disc on his first turn.
\item Find the probability that Tom wins the game on his second turn.
\item Find the probability that Sam removes a red disc on his first turn given that Tom wins the game on his second turn.\\

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2022 Q7 [8]}}