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Date November 2014 Marks available 3 Reference code 14N.1.hl.TZ0.4
Level HL only Paper 1 Time zone TZ0
Command term Determine Question number 4 Adapted from N/A

Question

Events A and B are such that P(A)=0.2 and P(B)=0.5.

Determine the value of P(AB) when

(i)     A and B are mutually exclusive;

(ii)     A and B are independent.

[4]
a.

Determine the range of possible values of P(A|B).

[3]
b.

Markscheme

(i)     use of P(AB)=P(A)+P(B)     (M1)

P(AB)=0.2+0.5

=0.7     A1

(ii)     use of P(AB)=P(A)+P(B)P(A)P(B)     (M1)

P(AB)=0.2+0.50.1

=0.6     A1

[4 marks]

a.

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

P(A|B) is a maximum when P(AB)=P(A)

P(A|B) is a minimum when P(AB)=0

0P(A|B)0.4     A1A1A1

 

Note:     A1 for each endpoint and A1 for the correct inequalities.

[3 marks]

Total [7 marks]

b.

Examiners report

This part was generally well done.

a.

Disappointingly, many candidates did not seem to understand the meaning of the word ‘range’ in this context.

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.4 » Conditional probability; the definition P(A|P)=P(AB)P(B) .

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