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Date November Example questions Marks available 6 Reference code EXN.2.AHL.TZ0.9
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 9 Adapted from N/A

Question

A biased coin is weighted such that the probability, p, of obtaining a tail is 0.6. The coin is tossed repeatedly and independently until a tail is obtained.

Let E be the event “obtaining the first tail on an even numbered toss”.

Find PE.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

METHOD 1

En is the event “the first tail occurs on the 2nd, 4th, 6th, …, 2nth toss”

PE=Σn=1PEn         (A1)

 

Note: Award A1 for deducing that either 1 head before a tail or 3 heads before a tail or 5 heads before a tail etc. is required. In other words, deduces 2n-1 heads before a tail.

 

PE=0.4×0.6+0.43×0.6+0.45×0.6+         M1A1

 

Note: Award M1 for attempting to form an infinite geometric series.

Note: Award A1 for PE=Σn=10.42n-10.6.

 

uses S=u11-r with u1=0.6×0.4 and r=0.42         (M1)

 

Note: Award M1 for using S=u11-r with u1=0.4 and r=0.42

 

=0.6×0.41-0.42        A1

=0.286 =27        A1

 

METHOD 2

let T1 be the event “tail occurs on the first toss”

uses PE=PET1PT1+PET1'PT1'         M1

concludes that PET1=0 and so PE=PET1'PT1'         R1

PET1'=PE'=1-PE         A1

 

Note: Award A1 for concluding: given that a tail is not obtained on the first toss, then PET1' is the probability that the first tail is obtained after a further odd number of tosses, PE'.

 

PT1'=0.4

PE=0.41-PE         A1

attempts to solve for PE         (M1)

=0.286 =27         A1

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

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