User interface language: English | Español

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)=PA+PB-P(AB)=0.68

substitution of PA·PB for P(AB) in P(AB)         (M1)

PA+PB-PAPB  =0.68

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

3PB+PB-3PBPB=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
Show 252 related questions
Topic 4—Statistics and probability

View options