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Date	May 2022	Marks available	5	Reference code	22M.1.SL.TZ2.9
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	9	Adapted from	N/A

Question

A biased four-sided die with faces labelled $1, 2, 3$ and $4$ is rolled and the result recorded. Let $X$ be the result obtained when the die is rolled. The probability distribution for $X$ is given in the following table where $p$ and $q$ are constants.

For this probability distribution, it is known that $E (X) = 2$ .

Nicky plays a game with this four-sided die. In this game she is allowed a maximum of five rolls. Her score is calculated by adding the results of each roll. Nicky wins the game if her score is at least ten.

After three rolls of the die, Nicky has a score of four.

David has two pairs of unbiased four-sided dice, a yellow pair and a red pair.

Both yellow dice have faces labelled $1, 2, 3$ and $4$ . Let $S$ represent the sum obtained by rolling the two yellow dice. The probability distribution for $S$ is shown below.

The first red die has faces labelled $1, 2, 2$ and $3$ . The second red die has faces labelled $1, a, a$ and $b$ , where $a < b$ and $a, b \in ℤ^{+}$ . The probability distribution for the sum obtained by rolling the red pair is the same as the distribution for the sum obtained by rolling the yellow pair.

Show that $p = 0.4$ and $q = 0.2$ .

[5]

Find $P (X > 2)$ .

[2]

Assuming that rolls of the die are independent, find the probability that Nicky wins the game.

[5]

Determine the value of $b$ .

[2]

Find the value of $a$ , providing evidence for your answer.

[2]

Markscheme

uses $\sum P (X = x) = 1$ to form a linear equation in $p$ and $q$ (M1)

correct equation in terms of $p$ and $q$ from summing to $1$ A1

$p + 0.3 + q + 0.1 = 1$ OR $p + q = 0.6$ (or equivalent)

uses $E (X) = 2$ to form a linear equation in $p$ and $q$ (M1)

correct equation in terms of $p$ and $q$ from $E (X) = 2$ A1

$p + 0.6 + 3 q + 0.4 = 2$ OR $p + 3 q = 1$ (or equivalent)

Note: The marks for using $\sum P (X = x) = 1$ and the marks for using $E (X) = 2$ may be awarded independently of each other.

evidence of correctly solving these equations simultaneously A1

for example, $2 q = 0.4 \Rightarrow q = 0.2$ or $p + 3 \times (0.6 - p) = 1 \Rightarrow p = 0.4$

so $p = 0.4$ and $q = 0.2$ AG

[5 marks]

valid approach (M1)

$P (X > 2) = P (X = 3) + P (X = 4)$ OR $P (X > 2) = 1 - P (X = 1) - P (X = 2)$

$= 0.3$ A1

[2 marks]

recognises at least one of the valid scores ( $6, 7$ , or $8$ ) required to win the game (M1)

Note: Award M0 if candidate also considers scores other than $6, 7$ , or $8$ (such as $5$ ).

let $T$ represent the score on the last two rolls

a score of $6$ is obtained by rolling $(2, 4), (4, 2)$ or $(3, 3)$

$P (T = 6) = 2 (0.3) (0.1) + {(0.2)}^{2} (= 0.1)$ A1

a score of $7$ is obtained by rolling $(3, 4)$ or $(4, 3)$

$P (T = 7) = 2 (0.2) (0.1) (= 0.04)$ A1

a score of $8$ is obtained by rolling $(4, 4)$

$P (T = 8) = {(0.1)}^{2} (= 0.01)$ A1

Note: The above 3 A1 marks are independent of each other.

$P (Nicky wins) = 0.1 + 0.04 + 0.01$

$= 0.15$ A1

[5 marks]

$3 + b = 8$ (M1)

$b = 5$ A1

[2 marks]

METHOD 1

EITHER

$P (S = 5) = \frac{4}{16}$

$P (S = a + 2) = \frac{4}{16}$ A1

$\Rightarrow a + 2 = 5$

$P (S = 6) = \frac{3}{16}$

$P (S = a + 3) = \frac{2}{16}$ and $P (S = 5 + 1) = \frac{1}{16}$ A1

$\Rightarrow a + 3 = 6$

$P (S = 4) = \frac{3}{16}$

$P (S = a + 1) = \frac{2}{16}$ and $P (S = 1 + 3) = \frac{1}{16}$ A1

$\Rightarrow a + 1 = 4$

THEN

$\Rightarrow a = 3$ A1

Note: Award A0A0 for $a = 3$ obtained without working/reasoning/justification.

METHOD 2

EITHER

correctly lists a relevant part of the sample space A1

for example, $\{S = 4\} = \{(3, 1), (1, a), (1, a)\}$ or $\{S = 5\} = \{(2, a), (2, a), (2, a), (2, a)\}$

or $\{S = 6\} = \{(3, a), (3, a), (1, 5)\}$

$a + 3 = 6$

eliminates possibilities (exhaustion) for $a < 5$

convincingly shows that $a \neq 2, 4$ A1

$a \neq 4$ , for example, $P (S = 7) = \frac{2}{16}$ from $(2, 5), (2, 5)$ and so

$(3, a), (3, a) \Rightarrow a + 3 \neq 7$

THEN

$\Rightarrow a = 3$ A1

[2 marks]

Examiners report

A majority of candidates knew to set up equations using the sum of the probabilities in the distribution equal to 1 and/or the expected value equal to 2, however some candidates simply substituted the given values of $p$ and $q$ into their equations, which is considered working backwards and not doing what is required by the command term "show that". For the candidates who did set up both equations in terms of $p$ and $q$ , nearly all were successful in solving the resulting system of equations. Many candidates answered part (b) correctly using the given values from part (a). In part (c), most candidates recognized a sum of 6 (or more) was required in the final two rolls, but very few were able to find all the different outcomes to make this happen, especially for sums that can happen in more than one way, such as $3 + 4$ and $4 + 3$ . While some candidates were able to correctly answer parts (d) and (e), some did not attempt
these questions parts, and many did not justify their final answer in part (e).

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.5—Probability concepts, expected numbers

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Topic 4—Statistics and probability » SL 4.6—Combined, mutually exclusive, conditional, independence, prob diagrams

Topic 4—Statistics and probability

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