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Date May 2022 Marks available 6 Reference code 22M.2.SL.TZ2.4
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 4 Adapted from N/A

Question

Events A and B are independent and P(A)=3P(B).

Given that P(AB)=0.68, find P(B).

Markscheme

P(AB)=P(A)+P(B)-P(AB)=0.68

substitution of P(A)·P(B) for P(AB) in P(AB)         (M1)

P(A)+P(B)-P(A)P(B)  (=0.68)

substitution of 3P(B) for P(A)         (M1)

3P(B)+P(B)-3P(B)P(B)=0.68  (or equivalent)         (A1)

 

Note: The first two M marks are independent of each other.

 

attempts to solve their quadratic equation         (M1)

P(B)=0.2, 1.133 (15, 1715)

P(B)=0.2 (=15)          A2

 

Note: Award A1 if both answers are given as final answers for P(B).

 

[6 marks]

Examiners report

This question proved difficult for many students. One common error was to use P(AB)=P(A)+P(B), which simplified the problem greatly, resulting in a linear, not a quadratic equation.

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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