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Date November 2017 Marks available 3 Reference code 17N.2.SL.TZ0.S_4
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_4 Adapted from N/A

Question

A discrete random variable X has the following probability distribution.

N17/5/MATME/SP2/ENG/TZ0/04

Find the value of k .

[4]
a.

Write down P ( X = 2 ) .

[1]
b.

Find P ( X = 2 | X > 0 ) .

[3]
c.

Markscheme

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

valid approach     (M1)

eg total probability = 1

correct equation     (A1)

eg 0.475 + 2 k 2 + k 10 + 6 k 2 = 1 ,   8 k 2 + 0.1 k 0.525 = 0

k = 0.25     A2     N3

[4 marks]

a.

P ( X = 2 ) = 0.025     A1     N1

[1 mark]

b.

valid approach for finding P ( X > 0 )     (M1)

eg 1 0.475 ,   2 ( 0.25 2 ) + 0.025 + 6 ( 0.25 2 ) ,   1 P ( X = 0 ) ,   2 k 2 + k 10 + 6 k 2

correct substitution into formula for conditional probability     (A1)

eg 0.025 1 0.475 ,   0.025 0.525

0.0476190

P ( X = 2 | X > 0 ) = 1 21 (exact), 0.0476     A1     N2

[3 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
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 » SL 4.7—Discrete random variables
Topic 4—Statistics and probability

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