IBDP Maths SL 4.1 Concepts of population, sample AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) [2 marks]
\( E(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{14}{6} = \frac{7}{3} \) (M1A1).
(b)(i) [3 marks]
Consider cases for total \( = 5 \): \( 1+1+3 \) and \( 1+2+2 \) (M1).
\( P(1,1,3) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} \), \( P(1,2,2) = \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} \).
\( 3 \times P(1,1,3) + 3 \times P(1,2,2) = 3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{72} \) (A1).
(b)(ii) [3 marks]
Median \( = 1 \) implies sequences \( 1,1,1 \), \( 1,1,2 \), or \( 1,1,3 \) (M1 for \( 1,1,1 \), M1 for \( 1,1,2 \) and \( 1,1,3 \)).
\( P(1,1,1) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \).
\( P(1,1,2) = \frac{1}{6} \times \frac{1}{6} \times \frac{2}{6} = \frac{2}{216} \), \( P(1,1,3) = \frac{1}{6} \times \frac{1}{6} \times \frac{3}{6} = \frac{3}{216} \).
\( P = \frac{1}{216} + 3 \times \frac{2}{216} + 3 \times \frac{3}{216} = \frac{16}{216} = \frac{2}{27} \) (A1).
(c) [3 marks]
Let \( X \) be the number of twos, \( X \sim \text{B}\left(10, \frac{1}{3}\right) \) (M1).
\( P(X < 4) = P(X \leq 3) = 0.559 \) (M1A1).
(d) [3 marks]
Let \( n \) be the number of balls drawn, \( P(\text{at least one 2}) = 1 – P(\text{no 2}) \) (M1).
\( 1 – \left(\frac{2}{3}\right)^n > 0.95 \Rightarrow \left(\frac{2}{3}\right)^n < 0.05 \) (M1).
\( n = 8 \) (A1).
(e) [8 marks]
Expected number of 1’s: \( 8 p_1 = 4.8 \Rightarrow p_1 = \frac{3}{5} \) (M1A1).
Variance of number of 2’s: \( 8 p_2 (1 – p_2) = 1.5 \) (M1).
\( p_2^2 – p_2 + \frac{1.5}{8} = 0 \Rightarrow p_2^2 – p_2 – 0.1875 = 0 \) (M1).
\( p_2 = \frac{1}{4} \) or \( \frac{3}{4} \) (A1).
Reject \( p_2 = \frac{3}{4} \) as \( p_1 + p_2 + p_3 = 1 \), and \( \frac{3}{5} + \frac{3}{4} > 1 \).
\( p_3 = 1 – \frac{3}{5} – \frac{1}{4} = \frac{3}{20} \) (A1).
LCM of denominators is \( 20 \), so minimum total number of balls is \( 20 \) (M1).
Number of 3’s = \( \frac{3}{20} \times 20 = 3 \) (A1).
Markscheme Answers:
(a) \( E(X) = \frac{7}{3} \) (M1A1)
(b)(i) \( \frac{7}{72} \) (M1A1)
(b)(ii) \( \frac{2}{27} \) (M1M1A1)
(c) \( 0.559 \) (M1M1A1)
(d) \( 8 \) (M1M1A1)
(e) \( 3 \) (M1A1M1M1A1A1M1A1)
Total [19 marks]
▶️ Answer/Explanation
(a) [3 marks]
\( P(X = 3) = (0.1)^3 = 0.001 \) (A1).
\( P(X = 4) = P(\text{VV}\bar{\text{V}}\text{V}) + P(\text{V}\bar{\text{V}}\text{VV}) + P(\bar{\text{V}}\text{VVV}) \) (M1).
\( = 3 \times (0.1)^3 \times 0.9 = 0.0027 \) (A1).
(b) [5 marks]
METHOD 1
Attempting to form equations in \( a \) and \( b \) (M1).
\( \frac{9 + 3a + b}{2000} = \frac{1}{1000} \Rightarrow 3a + b = -7 \) (A1).
\( \frac{16 + 4a + b}{2000} \times \frac{9}{10} = \frac{27}{10000} \Rightarrow 4a + b = -10 \) (A1).
Attempting to solve simultaneously (M1).
\( a = -3 \), \( b = 2 \) (A1).
METHOD 2
\( P(X = n) = \binom{n-1}{2} \times (0.1)^3 \times (0.9)^{n-3} \) (M1).
\( = \frac{(n-1)(n-2)}{2000} \times (0.9)^{n-3} \) (M1A1).
\( = \frac{n^2 – 3n + 2}{2000} \times (0.9)^{n-3} \) (A1).
\( a = -3 \), \( b = 2 \) (A1).
(c) [4 marks]
METHOD 1
EITHER
\( P(X = n) = \frac{n^2 – 3n + 2}{2000} \times (0.9)^{n-3} \) (M1).
OR
\( P(X = n) = \binom{n-1}{2} \times (0.1)^3 \times (0.9)^{n-3} \) (M1).
THEN
\( P(X = n) = \frac{(n-1)(n-2)}{2000} \times (0.9)^{n-3} \) (A1).
\( P(X = n-1) = \frac{(n-2)(n-3)}{2000} \times (0.9)^{n-4} \) (A1).
\( \frac{P(X = n)}{P(X = n-1)} = \frac{(n-1)(n-2)}{(n-2)(n-3)} \times 0.9 = \frac{0.9 (n-1)}{n-3} \) (A1).
METHOD 2
\( \frac{P(X = n)}{P(X = n-1)} = \frac{\frac{n^2 – 3n + 2}{2000} \times (0.9)^{n-3}}{\frac{(n-1)^2 – 3(n-1) + 2}{2000} \times (0.9)^{n-4}} \) (M1).
\( = \frac{0.9 (n^2 – 3n + 2)}{(n^2 – 5n + 6)} \) (A1A1).
\( = \frac{0.9 (n-1)(n-2)}{(n-2)(n-3)} = \frac{0.9 (n-1)}{n-3} \) (A1).
(d) [5 marks]
(i) Attempting to solve \( \frac{0.9 (n-1)}{n-3} = 1 \) for \( n \) (M1).
\( n = 21 \) (A1).
\( \frac{0.9 (n-1)}{n-3} < 1 \Rightarrow n > 21 \) (R1).
\( \frac{0.9 (n-1)}{n-3} > 1 \Rightarrow n < 21 \) (R1).
\( X \) has two modes (AG).
(ii) The modes are \( m_1 = 20 \) and \( m_2 = 21 \) (A1).
(e) [3 marks]
METHOD 1
\( Y \sim \text{B}(x, 0.1) \) (A1).
Attempting to solve \( P(Y \geq 3) > 0.5 \) (or \( 1 – P(Y \leq 2) > 0.5 \)) (M1).
\( x = 27 \) (A1).
METHOD 2
\( \sum_{n=0}^x P(X = n) > 0.5 \) (A1).
Attempting to solve for \( x \) (M1).
\( x = 27 \) (A1).
Markscheme Answers:
(a) \( P(X = 3) = 0.001 \), \( P(X = 4) = 0.0027 \) (A1M1A1)
(b) \( a = -3 \), \( b = 2 \) (M1A1A1M1A1)
(c) \( \frac{P(X = n)}{P(X = n-1)} = \frac{0.9 (n-1)}{n-3} \) (M1A1A1A1)
(d) (i) \( X \) has two modes; (ii) \( m_1 = 20 \), \( m_2 = 21 \) (M1A1R1R1A1)
(e) \( x = 27 \) (A1M1A1)
Total [20 marks]