IBDP Maths AHL 4.13 Use of Bayes’ theorem for a maximum of three events AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
METHOD 1 [6 marks]
Use a tree diagram or Bayes’ theorem to find \( P(I|L) \), where \( I \) is flying with IS Air, \( U \) is flying with UN Air, and \( L \) is luggage lost (M1).
Total flights: \( 70 + 65 = 135 \). So, \( P(I) = \frac{65}{135} \), \( P(U) = \frac{70}{135} \), \( P(L|I) = 0.23 \), \( P(L|U) = 0.18 \) (A1).
Apply Bayes’ theorem: \( P(I|L) = \frac{P(L|I)P(I)}{P(L|I)P(I) + P(L|U)P(U)} \) (M1).
Compute: \( P(I|L) = \frac{0.23 \times \frac{65}{135}}{0.23 \times \frac{65}{135} + 0.18 \times \frac{70}{135}} = \frac{\frac{1495}{13500}}{\frac{1495}{13500} + \frac{1260}{13500}} = \frac{1495}{1495 + 1260} = \frac{1495}{2755} = \frac{299}{551} \approx 0.543 \) (A1A1A1).
Answer: \( \frac{299}{551} \approx 0.543 \) (accept 0.542) (A1).
METHOD 2 [6 marks]
Calculate expected number of lost luggage cases per airline.
For UN Air: \( 0.18 \times 70 = 12.6 \) (M1A1).
For IS Air: \( 0.23 \times 65 = 14.95 \) (A1).
Total expected lost luggage: \( 12.6 + 14.95 = 27.55 \).
Probability: \( P(I|L) = \frac{\text{IS Air lost luggage}}{\text{Total lost luggage}} = \frac{14.95}{27.55} = \frac{1495}{2755} = \frac{299}{551} \approx 0.543 \) (M1A1).
Answer: \( 0.543 \) (A1).
Markscheme Answers:
Method 1: \( \frac{299}{551} \approx 0.543 \) (accept 0.542) (M1A1A1A1M1A1)
Method 2: \( 0.543 \) (M1A1A1M1A1A1)
Total [6 marks]
▶️ Answer/Explanation
EITHER (Tree Diagram Method)
Define events: \( C \) (car), \( B \) (bicycle), \( W \) (walking), \( L’ \) (on time). Given: \( P(C) = 0.3 \), \( P(B) = 0.2 \), \( P(W) = 0.5 \), \( P(L|C) = 0.05 \), \( P(L|B) = 0.1 \), \( P(L|W) = 0.25 \), so \( P(L’|C) = 0.95 \), \( P(L’|B) = 0.9 \), \( P(L’|W) = 0.75 \).
Tree diagram with first level probabilities: \( P(C) = 0.3 \), \( P(B) = 0.2 \), \( P(W) = 0.5 \) (M1A1). Second level: \( P(L’|C) = 0.95 \), \( P(L’|B) = 0.9 \), \( P(L’|W) = 0.75 \) (A1).
Total probability of being on time: \( P(L’) = P(L’|C)P(C) + P(L’|B)P(B) + P(L’|W)P(W) = 0.95 \times 0.3 + 0.9 \times 0.2 + 0.75 \times 0.5 = 0.285 + 0.18 + 0.375 = 0.84 \).
Probability: \( P(B|L’) = \frac{P(L’|B)P(B)}{P(L’)} = \frac{0.9 \times 0.2}{0.84} = \frac{0.18}{0.84} = \frac{18}{84} = \frac{3}{14} \approx 0.214 \) (M1A1A1).
OR (Bayes’ Theorem Method)
Use Bayes’ theorem: \( P(B|L’) = \frac{P(L’|B)P(B)}{P(L’|B)P(B) + P(L’|C)P(C) + P(L’|W)P(W)} \) (M1).
Given: \( P(B) = 0.2 \), \( P(L’|B) = 0.9 \), \( P(C) = 0.3 \), \( P(L’|C) = 0.95 \), \( P(W) = 0.5 \), \( P(L’|W) = 0.75 \) (A1A1).
Compute denominator: \( P(L’) = 0.9 \times 0.2 + 0.95 \times 0.3 + 0.75 \times 0.5 = 0.18 + 0.285 + 0.375 = 0.84 \) (M1).
Then: \( P(B|L’) = \frac{0.9 \times 0.2}{0.84} = \frac{0.18}{0.84} = \frac{18}{84} = \frac{3}{14} \approx 0.214 \) (A1A1).
Markscheme Answers:
\( \frac{3}{14} \approx 0.214 \) (M1A1A1M1A1A1)
Total [6 marks]