IBDP Maths SL 4.6 Venn and tree diagrams, counting principles AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Solution
(a) Calculating \( P(A \cap B) \):
\[ P(A) = \frac{4}{6} = \frac{2}{3} \]
After drawing one red ball, there are 5 balls left (3 red and 2 green):
\[ P(B|A) = \frac{2}{5} \]
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15} \]
✅ Answer: \(\frac{4}{15}\)
(b) Calculating \( P(C) \):
There are two ways to get exactly one red ball:
1. First ball red and second ball green: \( \frac{4}{6} \times \frac{2}{5} = \frac{4}{15} \)
2. First ball green and second ball red: \( \frac{2}{6} \times \frac{4}{5} = \frac{4}{15} \)
Total probability: \( \frac{4}{15} + \frac{4}{15} = \frac{8}{15} \)
✅ Answer: \(\frac{8}{15}\)
[1]▶️ Answer/Explanation
Solutions
Part (a)
Shade the region \( A’ \cap B’ \).
\( A’ \cap B’ \) represents the region outside both \( A \) and \( B \) in the Venn diagram.
Answer: Region outside both circles shaded.
Part (b)
Find \( P(A’|B’) \).
Conditional probability: \[ P(A’|B’) = \frac{P(A’ \cap B’)}{P(B’)} \]
Using De Morgan’s Law: \[ P(A’ \cap B’) = P((A \cup B)’) = 1 – P(A \cup B) \]
Given \( P(A \cup B) = 0.9 \), so: \[ P(A’ \cap B’) = 1 – 0.9 = 0.1 \]
Find \( P(B’) \): \[ P(B’) = P(A \cap B’) + P(A’ \cap B’) \]
Given \( P(A \cap B’) = 0.2 \), so: \[ P(B’) = 0.2 + 0.1 = 0.3 \]
\[ P(A’|B’) = \frac{0.1}{0.3} = \frac{1}{3} \]
Answer: \( P(A’|B’) = \frac{1}{3} \).