CAIE S1 2018 June — Question 2 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndependent Events
TypeReal-world independence interpretation
DifficultyModerate -0.3 This is a straightforward application of probability rules and independence testing. Part (i) requires P(A∪B) = P(A) + P(B) - P(A∩B) with clearly given proportions. Part (ii) tests independence by checking if P(M∩H) = P(M)×P(H), which is routine calculation. The context is clear, calculations are direct, and no problem-solving insight is needed—slightly easier than average due to its mechanical nature.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables
2 In a group of students, \(\frac { 3 } { 4 }\) are male. The proportion of male students who like their curry hot is \(\frac { 3 } { 5 }\) and the proportion of female students who like their curry hot is \(\frac { 4 } { 5 }\). One student is chosen at random.
  1. Find the probability that the student chosen is either female, or likes their curry hot, or is both female and likes their curry hot.
  2. Showing your working, determine whether the events 'the student chosen is male' and 'the student chosen likes their curry hot' are independent.
Question 2(i):
Method 1:
AnswerMarks Guidance
\(P(M \cap H) = \frac{3}{4} \times \frac{3}{5} = \frac{9}{20}\) (0.45)B1 Seen, accept unsimplified
\(P(F \text{ or } M \cap H) = \frac{1}{4} + \frac{9}{20} = \frac{14}{20}\)M1, A1 M1: Numerical attempt at \(P(F) + P(M \cap H)\); A1: Correct unsimplified expression
\(= \frac{7}{10}\) (0.7) OEA1 Correct final answer
Method 2:
AnswerMarks Guidance
\(P(M \cap H') = \frac{3}{4} \times \frac{2}{5} = \frac{6}{20}\) (0.3)B1 Seen, accept unsimplified
\(P(F \text{ or } M \cap H) = 1 - P(M \cap H')\)M1 Numerical attempt at \(1 - P(M \cap H')\)
\(= 1 - \frac{3}{4} \times \frac{2}{5}\)A1 Correct unsimplified expression
\(= \frac{7}{10}\) (0.7) OEA1 Correct final answer
Method 3:
AnswerMarks Guidance
\(P(F \cap H' \text{ or } H) = \frac{1}{4} \times \frac{1}{5} + \frac{1}{4} \times \frac{4}{5} + \frac{3}{4} \times \frac{3}{5}\)B1 \(\frac{3}{4} \times \frac{3}{5}\) (\(\frac{9}{20}\)) or \(\frac{1}{4} \times \frac{4}{5}\) (\(\frac{4}{20}\)) or \(\frac{3}{4} \times \frac{3}{5} + \frac{1}{4} \times \frac{4}{5}\) (\(\frac{13}{20}\)) seen
\(= \frac{1}{20} + \frac{4}{20} + \frac{9}{20}\)M1, A1 M1: Numerical attempt at \(P(F \cap H') + P(F \cap H) + P(M \cap H)\); A1: Correct unsimplified expression
\(= \frac{7}{10}\) (0.7)A1 Correct final answer
Method 4 – Venn diagram:
AnswerMarks Guidance
\(P(F \cup H) = P(F) + P(H) - P(F \cap H)\)B1 \(\frac{3}{4} \times \frac{3}{5}\) (\(\frac{9}{20}\)) or \(\frac{1}{4} \times \frac{4}{5}\) (\(\frac{4}{20}\)) or \(\frac{3}{4} \times \frac{3}{5} + \frac{1}{4} \times \frac{4}{5}\) (\(\frac{13}{20}\)) seen
\(= \frac{1}{4} + \frac{1}{4} \times \frac{4}{5} + \frac{3}{4} \times \frac{3}{5} - \frac{1}{4} \times \frac{4}{5}\)M1 Numerical attempt at \(P(F) + P(H) - P(F \cap H)\)
\(= \frac{1}{4} + \frac{4}{20} + \frac{9}{20} - \frac{4}{20}\)A1 Correct unsimplified expression
\(= \frac{7}{10}\) (0.7)A1 Correct final answer
Total: 4 marks
Question 2(ii):
Method 1:
AnswerMarks Guidance
\(P(M) \times P(H) = \frac{3}{4} \times \frac{13}{20} = \frac{39}{80}\)M1 Unsimplified, or better, legitimate numerical attempt at \(P(M) \times P(H)\) and \(P(M \cap H)\); descriptors \(P(M \cap H)\) and \(P(M) \times P(H)\) seen, correct numerical evaluation and comparison, conclusion stated
\(P(M \cap H) = \frac{3}{4} \times \frac{3}{5} = 0.45\)
AnswerMarks
\(\frac{39}{80}\) (0.4875) \(\neq\) 0.45, not independentA1
Method 2:
AnswerMarks Guidance
\(P(M \mid H) = \frac{P(M \cap H)}{P(H)} = \frac{\frac{9}{20}}{\frac{13}{20}} = \frac{9}{13}\)M1 Unsimplified, or better, numerical attempt at \(P(H)\) and \(P(M \cap H)\), \(P(M)\)
\(P(M) = \frac{3}{4}\)
AnswerMarks Guidance
\(\frac{9}{13} \neq \frac{3}{4}\), not independentA1 Descriptors \(P(M \cap H)\), \(P(H)\) and \(P(M)\) OR \(P(M
Total: 2 marks
## Question 2(i):

**Method 1:**

$P(M \cap H) = \frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$ (0.45) | B1 | Seen, accept unsimplified

$P(F \text{ or } M \cap H) = \frac{1}{4} + \frac{9}{20} = \frac{14}{20}$ | M1, A1 | M1: Numerical attempt at $P(F) + P(M \cap H)$; A1: Correct unsimplified expression

$= \frac{7}{10}$ (0.7) OE | A1 | Correct final answer

**Method 2:**

$P(M \cap H') = \frac{3}{4} \times \frac{2}{5} = \frac{6}{20}$ (0.3) | B1 | Seen, accept unsimplified

$P(F \text{ or } M \cap H) = 1 - P(M \cap H')$ | M1 | Numerical attempt at $1 - P(M \cap H')$

$= 1 - \frac{3}{4} \times \frac{2}{5}$ | A1 | Correct unsimplified expression

$= \frac{7}{10}$ (0.7) OE | A1 | Correct final answer

**Method 3:**

$P(F \cap H' \text{ or } H) = \frac{1}{4} \times \frac{1}{5} + \frac{1}{4} \times \frac{4}{5} + \frac{3}{4} \times \frac{3}{5}$ | B1 | $\frac{3}{4} \times \frac{3}{5}$ ($\frac{9}{20}$) or $\frac{1}{4} \times \frac{4}{5}$ ($\frac{4}{20}$) or $\frac{3}{4} \times \frac{3}{5} + \frac{1}{4} \times \frac{4}{5}$ ($\frac{13}{20}$) seen

$= \frac{1}{20} + \frac{4}{20} + \frac{9}{20}$ | M1, A1 | M1: Numerical attempt at $P(F \cap H') + P(F \cap H) + P(M \cap H)$; A1: Correct unsimplified expression

$= \frac{7}{10}$ (0.7) | A1 | Correct final answer

**Method 4 – Venn diagram:**

$P(F \cup H) = P(F) + P(H) - P(F \cap H)$ | B1 | $\frac{3}{4} \times \frac{3}{5}$ ($\frac{9}{20}$) or $\frac{1}{4} \times \frac{4}{5}$ ($\frac{4}{20}$) or $\frac{3}{4} \times \frac{3}{5} + \frac{1}{4} \times \frac{4}{5}$ ($\frac{13}{20}$) seen

$= \frac{1}{4} + \frac{1}{4} \times \frac{4}{5} + \frac{3}{4} \times \frac{3}{5} - \frac{1}{4} \times \frac{4}{5}$ | M1 | Numerical attempt at $P(F) + P(H) - P(F \cap H)$

$= \frac{1}{4} + \frac{4}{20} + \frac{9}{20} - \frac{4}{20}$ | A1 | Correct unsimplified expression

$= \frac{7}{10}$ (0.7) | A1 | Correct final answer

**Total: 4 marks**

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## Question 2(ii):

**Method 1:**

$P(M) \times P(H) = \frac{3}{4} \times \frac{13}{20} = \frac{39}{80}$ | M1 | Unsimplified, or better, legitimate numerical attempt at $P(M) \times P(H)$ and $P(M \cap H)$; descriptors $P(M \cap H)$ and $P(M) \times P(H)$ seen, correct numerical evaluation and comparison, conclusion stated

$P(M \cap H) = \frac{3}{4} \times \frac{3}{5} = 0.45$

$\frac{39}{80}$ (0.4875) $\neq$ 0.45, not independent | A1 |

**Method 2:**

$P(M \mid H) = \frac{P(M \cap H)}{P(H)} = \frac{\frac{9}{20}}{\frac{13}{20}} = \frac{9}{13}$ | M1 | Unsimplified, or better, numerical attempt at $P(H)$ and $P(M \cap H)$, $P(M)$

$P(M) = \frac{3}{4}$

$\frac{9}{13} \neq \frac{3}{4}$, not independent | A1 | Descriptors $P(M \cap H)$, $P(H)$ and $P(M)$ OR $P(M|H)$ and $P(M)$ seen, numerical evaluation and comparison, conclusion stated

**Total: 2 marks**

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2 In a group of students, $\frac { 3 } { 4 }$ are male. The proportion of male students who like their curry hot is $\frac { 3 } { 5 }$ and the proportion of female students who like their curry hot is $\frac { 4 } { 5 }$. One student is chosen at random.\\
(i) Find the probability that the student chosen is either female, or likes their curry hot, or is both female and likes their curry hot.\\

(ii) Showing your working, determine whether the events 'the student chosen is male' and 'the student chosen likes their curry hot' are independent.\\

\hfill \mbox{\textit{CAIE S1 2018 Q2 [6]}}
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