Question 4 - A-Level Maths

CAIE S1 2017 June — Question 4 6 marks

Exam Board	CAIE
Module	S1 (Statistics 1)
Year	2017
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Definitions
Type	Probability distribution finding parameters
Difficulty	Standard +0.3 This is a straightforward probability problem requiring students to set up equations from given conditions (score of 6 means both spinners show 3, score of 5 means one shows 2 and one shows 3) and solve the system p+q+r=1, r²=1/36, 2qr=1/9. The algebra is routine and the conceptual demand is modest—slightly easier than average for A-level.
Spec	2.03a Mutually exclusive and independent events 2.03b Probability diagrams: tree, Venn, sample space

4 Two identical biased triangular spinners with sides marked 1,2 and 3 are spun. For each spinner, the probabilities of landing on the sides marked 1,2 and 3 are \(p , q\) and \(r\) respectively. The score is the sum of the numbers on the sides on which the spinners land. You are given that \(\mathrm { P } (\) score is \(6 ) = \frac { 1 } { 36 }\) and \(\mathrm { P } (\) score is \(5 ) = \frac { 1 } { 9 }\). Find the values of \(p , q\) and \(r\).

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Question 4:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(P(\text{score is } 6) = P(3,3)\)	M1	Realising that score 6 is only \(P(3,3)\)
\(= r^2 = \frac{1}{36}\), \(r = \frac{1}{6}\)	A1	Correct ans [SR B2 \(r = \frac{1}{6}\) without workings]
\(P(2,3) + P(3,2) = \frac{1}{9}\); \(qr + rq = \frac{1}{9}\)	M1	Eqn involving \(qr\) (OE) equated to \(\frac{1}{9}\) (\(r\) may be replaced by their 'r value')
\(\frac{q}{6} + \frac{q}{6} = \frac{1}{9}\)	M1	Correct equation with their 'r value' substituted
\(q = \frac{1}{3}\)	A1	Correct answer seen, does not imply previous M's
\(p = 1 - \frac{1}{6} - \frac{1}{3} = \frac{1}{2}\)	B1 FT	FT their \(p\) + their \(r\) + their \(q = 1\), \(0 < p < 1\)
Total: 6

## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{score is } 6) = P(3,3)$ | M1 | Realising that score 6 is only $P(3,3)$ |
| $= r^2 = \frac{1}{36}$, $r = \frac{1}{6}$ | A1 | Correct ans [SR **B2** $r = \frac{1}{6}$ without workings] |
| $P(2,3) + P(3,2) = \frac{1}{9}$; $qr + rq = \frac{1}{9}$ | M1 | Eqn involving $qr$ (OE) equated to $\frac{1}{9}$ ($r$ may be replaced by *their* 'r value') |
| $\frac{q}{6} + \frac{q}{6} = \frac{1}{9}$ | M1 | Correct equation with *their* 'r value' substituted |
| $q = \frac{1}{3}$ | A1 | Correct answer seen, does **not** imply previous M's |
| $p = 1 - \frac{1}{6} - \frac{1}{3} = \frac{1}{2}$ | B1 FT | FT their $p$ + their $r$ + their $q = 1$, $0 < p < 1$ |
| **Total: 6** | | |

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4 Two identical biased triangular spinners with sides marked 1,2 and 3 are spun. For each spinner, the probabilities of landing on the sides marked 1,2 and 3 are $p , q$ and $r$ respectively. The score is the sum of the numbers on the sides on which the spinners land. You are given that $\mathrm { P } ($ score is $6 ) = \frac { 1 } { 36 }$ and $\mathrm { P } ($ score is $5 ) = \frac { 1 } { 9 }$. Find the values of $p , q$ and $r$.\\

\hfill \mbox{\textit{CAIE S1 2017 Q4 [6]}}

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Q1 5 Q2 6 Q3 6 Q4 6 Q5 9 Q6 9 Q7 9