CAIE S1 2006 June — Question 2 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2006
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTree Diagrams
TypeFind unknown probability parameter
DifficultyModerate -0.8 This is a straightforward tree diagram problem requiring basic probability rules: part (i) uses the law of total probability (0.2×0.75 + 0.8×x = 0.5) to find x, and part (ii) applies Bayes' theorem with values already computed. Both are standard textbook exercises with clear methods and minimal steps.
Spec2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles

2 The probability that Henk goes swimming on any day is 0.2 . On a day when he goes swimming, the probability that Henk has burgers for supper is 0.75 . On a day when he does not go swimming the probability that he has burgers for supper is \(x\). This information is shown on the following tree diagram. \includegraphics[max width=\textwidth, alt={}, center]{14e8a601-2180-4491-9336-cafd262f2596-2_693_1038_845_555} The probability that Henk has burgers for supper on any day is 0.5 .
  1. Find \(x\).
  2. Given that Henk has burgers for supper, find the probability that he went swimming that day.

Question 2(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(0.8\) seenB1
\(0.8x + (0.2)\times(0.75) = 0.5\)M1 Summing two 2-term brackets
M1Equating their LHS containing \(x\) to \(0.5\)
\(x = 0.438\)A1 4 Correct answer
Question 2(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(S \text{ given } B) = 0.15/0.5\)M1 Correct numerator
\(= 0.3\)A1 2 Correct answer
## Question 2(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.8$ seen | B1 | |
| $0.8x + (0.2)\times(0.75) = 0.5$ | M1 | Summing two 2-term brackets |
| | M1 | Equating their LHS containing $x$ to $0.5$ |
| $x = 0.438$ | A1 | **4** Correct answer |

## Question 2(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(S \text{ given } B) = 0.15/0.5$ | M1 | Correct numerator |
| $= 0.3$ | A1 | **2** Correct answer |

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2 The probability that Henk goes swimming on any day is 0.2 . On a day when he goes swimming, the probability that Henk has burgers for supper is 0.75 . On a day when he does not go swimming the probability that he has burgers for supper is $x$. This information is shown on the following tree diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{14e8a601-2180-4491-9336-cafd262f2596-2_693_1038_845_555}

The probability that Henk has burgers for supper on any day is 0.5 .\\
(i) Find $x$.\\
(ii) Given that Henk has burgers for supper, find the probability that he went swimming that day.

\hfill \mbox{\textit{CAIE S1 2006 Q2 [6]}}