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.

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]

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