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Date None Specimen Marks available 6 Reference code SPNone.1.sl.TZ0.9
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 9 Adapted from N/A

Question

Two standard six-sided dice are tossed. A diagram representing the sample space is shown below.


Let \(X\) be the sum of the scores on the two dice.

(i)     Find \({\rm{P}}(X = 6)\) .

(ii)    Find \({\rm{P}}(X > 6)\) .

(iii)   Find \({\rm{P}}(X = 7|X > 6)\) .

[6]
a(i), (ii) and (iii).

Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which the game is fair.

[8]
b.

Markscheme

(i) number of ways of getting \(X = 6\) is 5     A1

\({\rm{P}}(X = 6) = \frac{5}{{36}}\)     A1     N2

(ii) number of ways of getting \(X > 6\) is 21     A1

\({\rm{P}}(X > 6) = \frac{{21}}{{36}}\left( { = \frac{7}{{12}}} \right)\)     A1     N2

(iii) \({\rm{P}}(X = 7|X > 6) = \frac{6}{{21}}\left( { = \frac{2}{7}} \right)\)     A2     N2

[6 marks]

a(i), (ii) and (iii).

attempt to find \({\rm{P}}(X < 6)\)     M1

e.g. \(1 - \frac{5}{{36}} - \frac{{21}}{{36}}\)

\({\rm{P}}(X < 6) = \frac{{10}}{{36}}\)     A1

fair game if \({\rm{E}}(W) = 0\) (may be seen anywhere)     R1

attempt to substitute into \({\rm{E}}(X)\) formula     M1

e.g. \(3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) - k\left( {\frac{{10}}{{36}}} \right)\)

correct substitution into \({\rm{E}}(W) = 0\)     A1

e.g. \(3\left( {\frac{5}{{36}}} \right) + 1\left( {\frac{{21}}{{36}}} \right) - k\left( {\frac{{10}}{{36}}} \right) = 0\)

work towards solving     M1

e.g. \(15 + 21 - 10k = 0\)

\(36 = 10k\)     A1

\(k = \frac{{36}}{{10}}( = 3.6)\)     A1     N4

[8 marks]

b.

Examiners report

[N/A]
a(i), (ii) and (iii).
[N/A]
b.

Syllabus sections

Topic 5 - Statistics and probability » 5.5 » The probability of an event \(A\) is \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( U \right)}}\) .
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