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Date	May 2021	Marks available	2	Reference code	21M.1.SL.TZ1.9
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find and Hence	Question number	9	Adapted from	N/A

Question

A biased four-sided die, $A$ , is rolled. Let $X$ be the score obtained when die $A$ is rolled. The probability distribution for $X$ is given in the following table.

A second biased four-sided die, $B$ , is rolled. Let $Y$ be the score obtained when die $B$ is rolled.
The probability distribution for $Y$ is given in the following table.

Find the value of $p$ .

[2]

Hence, find the value of $E (X)$ .

[2]

State the range of possible values of $r$ .

[1]

c.i.

Hence, find the range of possible values of $q$ .

[2]

c.ii.

Hence, find the range of possible values for $E (Y)$ .

[3]

Agnes and Barbara play a game using these dice. Agnes rolls die $A$ once and Barbara rolls die $B$ once. The probability that Agnes’ score is less than Barbara’s score is $\frac{1}{2}$ .

Find the value of $E (Y)$ .

[6]

Markscheme

recognising probabilities sum to $1$ (M1)

$p + p + p + \frac{1}{2} p = 1$

$p = \frac{2}{7}$ A1

[2 marks]

valid attempt to find $E (X)$ (M1)

$1 \times p + 2 \times p + 3 \times p + 4 \times \frac{1}{2} p (= 8 p)$

$E (X) = \frac{16}{7}$ A1

[2 marks]

$0 \leq r \leq 1$ A1

[1 mark]

c.i.

attempt to find a value of $q$ (M1)

$0 \leq 1 - 3 q \leq 1$ OR $r = 0 \Rightarrow q = \frac{1}{3}$ OR $r = 1 \Rightarrow q = 0$

$0 \leq q \leq \frac{1}{3}$ A1

[2 marks]

c.ii.

$E (Y) = 1 \times q + 2 \times q + 3 \times q + 4 \times r$ ( $= 2 + 2 r$ OR $4 - 6 q$ ) (A1)

one correct boundary value A1

$1 \times \frac{1}{3} + 2 \times \frac{1}{3} + 3 \times \frac{1}{3} + 4 \times 0 (= 2)$ OR

$1 \times 0 + 2 \times 0 + 3 \times 0 + 4 \times 1 (= 4)$ OR

$2 + 2 (0) (= 2)$ OR

$2 + 2 (1) (= 4)$ OR

$4 - 6 (0) (= 4)$ OR $4 - 6 (\frac{1}{3}) (= 2)$

$2 \leq E (Y) \leq 4$ A1

[3 marks]

METHOD 1

evidence of choosing at least four correct outcomes from

$1 & 2, 1 & 3, 1 & 4, 2 & 3, 2 & 4, 3 & 4$ (M1)

$\frac{6}{7} q + \frac{6}{7} r$ OR $3 p q + 3 p r$ OR $p q + p q + p (1 - 3 q) + p q + p (1 - 3 q) + p (1 - 3 q)$ (A1)

solving for either $q$ or $r$ M1

$\frac{6}{7} (q + 1 - 3 q) = \frac{1}{2}$ OR $\frac{6}{7} (\frac{1 - r}{3} + r) = \frac{1}{2}$ OR $3 p q + 3 p (1 - 3 q) = \frac{1}{2}$

OR $3 p (\frac{1 - r}{3}) + 3 p r = \frac{1}{2}$

EITHER two correct values

$q = \frac{5}{24}$ and $r = \frac{3}{8}$ A1A1

OR one correct value

$q = \frac{5}{24}$ OR $r = \frac{3}{8}$ A1

substituting their value for $q$ or $r$ A1

$4 - 6 (\frac{5}{24})$ OR $2 + 2 (\frac{3}{8})$

THEN

$E (Y) = \frac{11}{4}$ A1

METHOD 2 (solving for $E (Y)$ )

evidence of choosing at least four correct outcomes from

$1 & 2, 1 & 3, 1 & 4, 2 & 3, 2 & 4, 3 & 4$ (M1)

$\frac{6}{7} q + \frac{6}{7} r$ OR $3 p q + 3 p r$ OR $p q + p q + p (1 - 3 q) + p q + p (1 - 3 q) + p (1 - 3 q)$ (A1)

rearranging to make $q$ the subject M1

$q = \frac{4 - E (Y)}{6}$

$3 p q + 3 p (1 - 3 q) = \frac{1}{2}$ M1

$\frac{6}{7} \times (\frac{4 - E (Y)}{6}) + \frac{6}{7} (1 - 3 (\frac{4 - E (Y)}{6})) = \frac{1}{2}$ A1

$\frac{2 (E (Y) - 1)}{7} = \frac{1}{2}$

$E (Y) = \frac{11}{4}$ A1

[6 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.7—Discrete random variables

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Find the value of $E (Y)$ .
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medium.

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Topic 4—Statistics and probability

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