Home / IBDP Maths AA: Topic: AHL 4.13: Use of Bayes’ theorem for a maximum of three events: IB style Questions HL Paper 3

IBDP Maths AA: Topic: AHL 4.13: Use of Bayes’ theorem for a maximum of three events: IB style Questions HL Paper 3

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.

▶️Answer/Explanation

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.

Question

(a)     Find the percentage of the population that has been vaccinated.

(b)     A randomly chosen person catches the virus. Find the probability that this person has been vaccinated.

▶️Answer/Explanation

Markscheme

(a)

using the law of total probabilities:     (M1)

\(0.1p + 0.3\left( {1 – p} \right) = 0.22\)     A1

\(0.1p + 0.3 – 0.3p = 0.22\)

\(0.2p = 0.88\)

\(p = \frac{{0.88}}{{0.2}} = 0.4\)

\(p = 40\% \) (accept \(0.4\))     A1

(b)     required probability \( = \frac{{0.4 \times 0.1}}{{0.22}}\)     M1

\( = \frac{2}{{11}}\)   (\(0.182\))     A1

[5 marks]

Examiners report

Most candidates who successfully answered this question had first drawn a tree diagram, using a symbol to denote the probability that a randomly chosen person had received the influenza virus. For those who did not draw a tree diagram, there was poor understanding of how to apply the conditional probability formula.

Question

Natasha lives in Chicago and has relatives in Nashville and St. Louis.

Each time she visits her relatives, she either flies or drives.

When travelling to Nashville, the probability that she drives is \(\frac{4}{5}\), and when travelling to St. Louis, the probability that she flies is \(\frac{1}{3}\).

Given that the probability that she drives when visiting her relatives is \(\frac{13}{18}\), find the probability that for a particular trip,

a.she travels to Nashville;[3]

b.she is on her way to Nashville, given that she is flying.[3]

▶️Answer/Explanation

Markscheme

attempt to set up the problem using a tree diagram and/or an equation, with the unknown \(x\)     M1

\(\frac{4}{5}x + \frac{2}{3}(1 – x) = \frac{{13}}{{18}}\)     A1

\(\frac{{4x}}{5} – \frac{{2x}}{3} = \frac{{13}}{{18}} – \frac{2}{3}\)

\(\frac{{2x}}{{15}} = \frac{1}{{18}}\)

\(x = \frac{5}{{12}}\)     A1

[3 marks]

a.

attempt to set up the problem using conditional probability     M1

EITHER

\(\frac{{\frac{5}{{12}} \times \frac{1}{5}}}{{1 – \frac{{13}}{{18}}}}\)     A1

OR

\(\frac{{\frac{5}{{12}} \times \frac{1}{5}}}{{\frac{1}{{12}} + \frac{7}{{36}}}}\)     A1

THEN

\( = \frac{3}{{10}}\)     A1

[3 marks]

Total [6 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

Josie has three ways of getting to school. \(30\% \) of the time she travels by car, \(20\% \) of the time she rides her bicycle and \(50\% \) of the time she walks.

When travelling by car, Josie is late \(5\% \) of the time. When riding her bicycle she is late \(10\% \) of the time. When walking she is late \(25\% \) of the time. Given that she was on time, find the probability that she rides her bicycle.

▶️Answer/Explanation

Markscheme

EITHER

     M1A1A1

Note:     Award M1 for a two-level tree diagram, A1 for correct first level probabilities, and A1 for correct second level probabilities.

OR

\({\text{P}}(B|L’) = \frac{{{\text{P}}(L’|B){\text{P}}(B)}}{{{\text{P}}(L’|B){\text{P}}(B) + {\text{P}}(L’|C){\text{P}}(C) + {\text{P}}(L’|W){\text{P}}(Q)}}\;\;\;\left( { = \frac{{{\text{P}}(B \cap L’}}{{{\text{P}}(L’)}}} \right)\)     (M1)(A1)(A1)

THEN

\({\text{P}}(B|L’) = \frac{{0.9 \times 0.2}}{{0.9 \times 0.2 + 0.95 \times 0.3 + 0.75 \times 0.5}}\;\;\;\left( { = \frac{{0.18}}{{0.84}}} \right)\)     M1A1

\( = 0.214\;\;\;\left( { = \frac{3}{{14}}} \right)\)     A1

[6 marks]

Examiners report

[N/A]
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