CAIE S1 2021 November — Question 3 5 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2021
SessionNovember
Marks5
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TopicConditional Probability
TypeBayes with complementary outcome
DifficultyModerate -0.8 This is a straightforward application of Bayes' theorem with clearly stated probabilities and a simple tree diagram structure. Students need to find P(no biscuit), then apply the conditional probability formula, but all values are given directly with no algebraic manipulation or conceptual subtlety required.
Spec2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles
3 For her bedtime drink, Suki has either chocolate, tea or milk with probabilities \(0.45,0.35\) and 0.2 respectively. When she has chocolate, the probability that she has a biscuit is 0.3 When she has tea, the probability that she has a biscuit is 0.6 . When she has milk, she never has a biscuit. Find the probability that Suki has tea given that she does not have a biscuit.
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(T\B') = \frac{P(T \cap B')}{P(B')}\) M1
\(P(B') = 0.45 \times 0.7 + 0.35 \times 0.4 + 0.2 \times 1 = 0.655, \frac{131}{200}\)A1 Correct, accept unsimplified
\(P(T \cap B') = 0.35 \times 0.4 \left[= 0.14, \frac{7}{50}\right]\)M1 Seen as numerator or denominator of a fraction
\(P(T \B') = \frac{\textit{their } 0.14}{\textit{their } 0.655}\) M1
\(0.214, \frac{28}{131}\)A1 If 0 marks awarded, SC B1 0.214 WWW 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(T\|B') = \frac{P(T \cap B')}{P(B')}$ | M1 | $0.45 \times a + 0.35 \times b + 0.2[\times 1]$, $a = 0.7, 0.3b = 0.4, 0.6$, seen anywhere |
| $P(B') = 0.45 \times 0.7 + 0.35 \times 0.4 + 0.2 \times 1 = 0.655, \frac{131}{200}$ | A1 | Correct, accept unsimplified |
| $P(T \cap B') = 0.35 \times 0.4 \left[= 0.14, \frac{7}{50}\right]$ | M1 | Seen as numerator or denominator of a fraction |
| $P(T \| B') = \frac{\textit{their } 0.14}{\textit{their } 0.655}$ | M1 | Values substituted into conditional probability formula correctly. Accept unsimplified. Denominator sum of 3 two-factor probabilities (condone omission of 1 from final factor). If clearly identified, condone from incomplete denominator |
| $0.214, \frac{28}{131}$ | A1 | If 0 marks awarded, **SC B1** 0.214 WWW |

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3 For her bedtime drink, Suki has either chocolate, tea or milk with probabilities $0.45,0.35$ and 0.2 respectively. When she has chocolate, the probability that she has a biscuit is 0.3 When she has tea, the probability that she has a biscuit is 0.6 . When she has milk, she never has a biscuit.

Find the probability that Suki has tea given that she does not have a biscuit.\\

\hfill \mbox{\textit{CAIE S1 2021 Q3 [5]}}
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