IBDP Maths SL 4.6 Venn and tree diagrams, counting principles AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) [2 marks]
Use the combined events formula (M1):
\( P(S \cup T) = P(S) + P(T) – P(S \cap T) \).
\( P(S’ \cap T’) = 0.18 \), so \( P(S \cup T) = 1 – 0.18 = 0.82 \).
\( 0.82 = 0.7 + 0.2 – P(S \cap T) \).
\( P(S \cap T) = 0.7 + 0.2 – 0.82 = 0.08 \) (A1).
(b) [2 marks]
\( P(T \cap S’) = P(T) – P(S \cap T) \) (M1).
\( P(T \cap S’) = 0.2 – 0.08 = 0.12 \) (A1).
(c) [2 marks]
\( P(G \cap T) = P(G) \times P(T | G) \) (M1).
\( P(G \cap T) = 0.48 \times 0.25 = 0.12 \) (A1).
(d) [2 marks]
For independence, check if \( P(G \cap T) = P(G) \times P(T) \) (M1).
\( P(G) \times P(T) = 0.48 \times 0.2 = 0.096 \).
Since \( 0.096 \neq 0.12 = P(G \cap T) \), events \( G \) and \( T \) are not independent (A1).
Markscheme Answers:
(a) \( P(S \cap T) = 0.08 \) (M1A1)
(b) \( P(T \cap S’) = 0.12 \) (M1A1)
(c) \( P(G \cap T) = 0.12 \) (M1A1)
(d) Events \( G \) and \( T \) are not independent, as \( P(G) \times P(T) = 0.096 \neq 0.12 = P(G \cap T) \) (M1A1)
Total [8 marks]
▶️ Answer/Explanation
Part A.
(a) [2 marks]
\( X \sim \text{N}(231, 1.5^2) \) (M1).
\( P(X < 228) = 0.0228 \) (A1).
Note: Accept \( 0.0227 \).
(b)(i) [3 marks]
\( X \sim \text{N}(\mu, 1.5^2) \) (M1).
\( P(X < 228) = 0.002 \).
\( \frac{228 – \mu}{1.5} = -2.878\ldots \) (A1).
\( \mu = 232 \) grams (A1).
(b)(ii) [3 marks]
\( X \sim \text{N}(231, \sigma^2) \) (M1).
\( \frac{228 – 231}{\sigma} = -2.878\ldots \) (A1).
\( \sigma = 1.04 \) grams (A1).
(c) [3 marks]
\( X \sim \text{B}(100, 0.002) \) (M1).
\( P(X \leq 1) = 0.982\ldots \) (A1).
\( P(X \geq 2) = 1 – P(X \leq 1) = 0.0174 \) (A1).
Total [11 marks]
Part B.
(a) [3 marks]
Boys can be chosen in \( \frac{6 \times 5}{2} = 15 \) ways (A1).
Girls can be chosen in \( \frac{5 \times 4}{2} = 10 \) ways (A1).
Total \( = 15 \times 10 = 150 \) ways (A1).
(b) [2 marks]
Number of ways \( = 5 \times 4 = 20 \) (M1A1).
(c) [1 mark]
\( \frac{20}{150} = \frac{2}{15} \) (A1).
(d) [4 marks]
METHOD 1
\( P(T) = \frac{1}{5} \), \( P(A) = \frac{2}{5} \) (A1).
\( P(\text{T or A but not both}) = P(T) \times P(A’) + P(T’) \times P(A) \) (M1A1).
\( = \frac{1}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{2}{5} = \frac{11}{25} \) (A1).
METHOD 2
Number of selections including Fred \( = 5 \times \binom{5}{2} = 50 \) (A1).
Number of selections including Tim but not Anna \( = \binom{4}{2} = 6 \) (A1).
Number of selections including Anna but not Tim \( = 4 \times 4 = 16 \) (A1).
\( P(\text{T or A but not both}) = \frac{6 + 16}{50} = \frac{11}{25} \) (M1).
Total [10 marks]
Markscheme Answers:
Part A.
(a) \( P(X < 228) = 0.0228 \) (M1A1)
(b)(i) \( \mu = 232 \) grams (M1A1A1)
(b)(ii) \( \sigma = 1.04 \) grams (M1A1A1)
(c) \( P(X \geq 2) = 0.0174 \) (M1A1A1)
Part B.
(a) \( 150 \) ways (A1A1A1)
(b) \( 20 \) ways (M1A1)
(c) \( \frac{2}{15} \) (A1)
(d) \( \frac{11}{25} \) (A1M1A1A1)
Total [21 marks]