IBDP Maths AHL 4.13 Use of Bayes’ theorem for a maximum of three events AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Let \( x \) be the probability she travels to Nashville. Then \( 1 – x \) is the probability she travels to St. Louis.
Probability she drives = \( \frac{4}{5}x + \frac{2}{3}(1 – x) = \frac{13}{18} \).
Solve: \( \frac{4}{5}x + \frac{2}{3} – \frac{2}{3}x = \frac{13}{18} \).
Common denominator: \( \frac{12}{15}x + \frac{10}{15} – \frac{10}{15}x = \frac{13}{18} \).
\( \frac{2}{15}x + \frac{10}{15} = \frac{13}{18} \).
\( \frac{2}{15}x = \frac{13}{18} – \frac{10}{15} = \frac{13}{18} – \frac{12}{18} = \frac{1}{18} \).
\( x = \frac{1}{18} \cdot \frac{15}{2} = \frac{15}{36} = \frac{5}{12} \).
\(\boxed{\frac{5}{12}}\)
Probability she flies to Nashville = \( \frac{5}{12} \cdot \frac{1}{5} = \frac{1}{12} \).
Probability she flies to St. Louis = \( \frac{7}{12} \cdot \frac{1}{3} = \frac{7}{36} \).
Total probability she flies = \( \frac{1}{12} + \frac{7}{36} = \frac{3}{36} + \frac{7}{36} = \frac{10}{36} = \frac{5}{18} \).
Conditional probability = \( \frac{\frac{1}{12}}{\frac{5}{18}} = \frac{1}{12} \cdot \frac{18}{5} = \frac{18}{60} = \frac{3}{10} \).
\(\boxed{\frac{3}{10}}\)
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]
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]