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Date November 2016 Marks available 6 Reference code 16N.1.AHL.TZ0.H_10
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Show that and Hence Question number H_10 Adapted from N/A

Question

Consider two events A and A defined in the same sample space.

Given that P ( A B ) = 4 9 ,  P ( B | A ) = 1 3  and P ( B | A ) = 1 6 ,

Show that P ( A B ) = P ( A ) + P ( A B ) .

[3]
a.

(i)     show that P ( A ) = 1 3 ;

(ii)     hence find P ( B ) .

[6]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

P ( A B ) = P ( A ) + P ( B ) P ( A B )    M1

= P ( A ) + P ( A B ) + P ( A B ) P ( A B )    M1A1

= P ( A ) + P ( A B )    AG

METHOD 2

P ( A B ) = P ( A ) + P ( B ) P ( A B )    M1

= P ( A ) + P ( B ) P ( A | B ) × P ( B )    M1

= P ( A ) + ( 1 P ( A | B ) ) × P ( B )

= P ( A ) + P ( A | B ) × P ( B )    A1

= P ( A ) + P ( A B )    AG

[3 marks]

a.

(i)     use P ( A B ) = P ( A ) + P ( A B ) and P ( A B ) = P ( B | A ) P ( A )      (M1)

4 9 = P ( A ) + 1 6 ( 1 P ( A ) )    A1

8 = 18 P ( A ) + 3 ( 1 P ( A ) )    M1

P ( A ) = 1 3    AG

(ii)     METHOD 1

P ( B ) = P ( A B ) + P ( A B )    M1

= P ( B | A ) P ( A ) + P ( B | A ) P ( A )    M1

= 1 3 × 1 3 + 1 6 × 2 3 = 2 9    A1

METHOD 2

P ( A B ) = P ( B | A ) P ( A ) P ( A B ) = 1 3 × 1 3 = 1 9    M1

P ( B ) = P ( A B ) + P ( A B ) P ( A )    M1

P ( B ) = 4 9 + 1 9 1 3 = 2 9    A1

[6 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

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

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