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Date November 2017 Marks available 2 Reference code 17N.2.AHL.TZ0.H_2
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Hence and Show that Question number H_2 Adapted from N/A

Question

Events A and B are such that P ( A B ) = 0.95 ,  P ( A B ) = 0.6 and P ( A | B ) = 0.75 .

Find  P ( B ) .

[2]
a.

Find P ( A ) .

[2]
b.

Hence show that events A and B are independent.

[2]
c.

Markscheme

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

P ( A | B ) = P ( A B ) P ( B )

0.75 = 0.6 P ( B )     (M1)

P ( B )   ( = 0.6 0.75 ) = 0.8     A1

[2 marks]

a.

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

0.95 = P ( A ) + 0.8 0.6     (M1)

P ( A ) = 0.75     A1

[2 marks]

b.

METHOD 1

P ( A | B ) = P ( A B ) P ( B ) = 0.2 0.8 = 0.25     A1

P ( A | B ) = P ( A )     R1

hence A and B are independent     AG

 

Note:     If there is evidence that the student has calculated P ( A B ) = 0.2 by assuming independence in the first place, award A0R0.

 

METHOD 2

EITHER

P ( A ) = P ( A | B )     A1

OR

P ( A ) × P ( B ) = 0.75 × 0.80 = 0.6 = P ( A B )     A1

THEN

A and B are independent     R1

hence A and B are independent     AG

METHOD 3

P ( A ) × P ( B ) = 0.25 × 0.80 = 0.2     A1

P ( A ) × P ( B ) = P ( A B )     R1

hence A and B are independent     AG

[2 marks]

c.

Examiners report

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

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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