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Date May 2008 Marks available 6 Reference code 08M.2.hl.TZ1.8
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 8 Adapted from N/A

Question

Only two international airlines fly daily into an airport. UN Air has 70 flights a day and IS Air has 65 flights a day. Passengers flying with UN Air have an 18 % probability of losing their luggage and passengers flying with IS Air have a 23 % probability of losing their luggage. You overhear someone in the airport complain about her luggage being lost.

Find the probability that she travelled with IS Air.

Markscheme

METHOD 1

     (M1)

Let P(I) be the probability of flying IS Air, P(U) be the probability flying UN Air and P(L) be the probability of luggage lost.

\({\text{P}}(I|L) = \frac{{{\text{P}}(I \cap L)}}{{{\text{P}}(L)}}{\text{ }}\left( {{\text{or Bayes' formula , P}}(I|L) = \frac{{{\text{P}}(L|I){\text{P}}(I)}}{{{\text{P}}(L|I){\text{P}}(I) + {\text{P}}(L|U){\text{P}}(U)}}} \right)\)     (M1)

\( = \frac{{0.23 \times \frac{{65}}{{135}}}}{{0.18 \times \frac{{70}}{{135}} + 0.23 \times \frac{{65}}{{135}}}}\)     A1A1A1

\( = \frac{{299}}{{551}}{\text{ }}( = 0.543,{\text{ accept }}0.542)\)     A1

[6 marks] 

METHOD 2

Expected number of suitcases lost by UN Air is \(0.18 \times 70 = 12.6\)     M1A1

Expected number of suitcases lost by IS Air is \(0.23 \times 65 = 14.95\)     A1

\({\text{P}}(I|L) = \frac{{14.95}}{{12.6 + 14.95}}\)     M1A1

\( = 0.543\)     A1

[6 marks]

Examiners report

This question was well answered by the majority of candidates. Most candidates used either tree diagrams or expected value methods.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.4 » Use of Bayes’ theorem for a maximum of three events.

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