IB Mathematics AHL 4.14 Variance of random variable AA HL Paper 3- Exam Style Questions- New Syllabus
Question
This question explores the use of generating functions to analyze discrete probability distributions.
Two unbiased tetrahedral (four-sided) dice with faces labelled 1, 2, 3 and 4 are thrown and the scores recorded. The random variable \( M \) denotes the maximum of these two scores. The probability distribution of \( M \) is given in the following table.
| \( m \) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| \( P(M = m) \) | \( \frac{1}{16} \) | \( \frac{3}{16} \) | \( \frac{5}{16} \) | \( \frac{7}{16} \) |
(a) Calculate the expected value \( E(M) \).
The probability distribution of \( M \) can be represented by a probability generating function, \( G \), defined as \( G(t) = \sum_{m=1}^{4} P(M = m) t^{m} \). For this specific distribution, \( G(t) = \frac{1}{16} t + \frac{3}{16} t^2 + \frac{5}{16} t^3 + \frac{7}{16} t^4 \).
(b) Determine the value of \( G(1) \).
(c) (i) Differentiate the expression to find \( G'(t) \).
(ii) Hence, demonstrate that \( G'(1) = E(M) \).
A bag contains two red balls and three yellow balls. Two balls are selected at random without replacement. The random variable \( X \) represents the total number of red balls selected. The probability distribution of \( X \) is modeled by the function
\[ G_X(t) = \sum_{x=0}^{2} P(X=x)t^x. \]
(d) Show that \( G_X(t) = \frac{3}{10} + \frac{3}{5}t + \frac{1}{10}t^2 \), justifying the coefficients through probability calculations.
An unbiased coin and a biased coin are tossed simultaneously. The probability of the biased coin landing on tails is \( p \). Let \( Y \) be the random variable for the total number of tails obtained. The distribution of \( Y \) is given by
\[ G_Y(t) = \sum_{y=0}^{2} P(Y=y)t^y. \ ]
(e) Given that the coefficient of \( t^2 \) in \( G_Y(t) \) is \( \frac{1}{3} \), determine:
(i) the value of \( p \);
(ii) the full expression for \( G_Y(t) \).
Consider a new random variable \( Z = X + Y \), which represents the sum of the red balls and the tails obtained. Its distribution is given by the product
\[ G_z(t) = G_X(t)G_Y(t). \]
(f) Using the property that \( G_z'(1) = E(Z) \), find the value of \( E(Z) \).
Most-appropriate topic codes (IB Mathematics: Analysis and Approaches AHL):
• AHL 2.12: Polynomial functions, their graphs and equations — Parts (c), (d)
• AHL 1.15: Use of formal logic and demonstration to justify mathematical results — Parts (c)(ii), (d)
▶️ Answer/Explanation
(a)
Method:
\( E(M) = \sum m P(M = m) \)
\( = 1 \times \frac{1}{16} + 2 \times \frac{3}{16} + 3 \times \frac{5}{16} + 4 \times \frac{7}{16} \)
\( = \frac{1+6+15+28}{16} = \frac{50}{16} = \frac{25}{8} = 3.125 \)
\( \boxed{\frac{25}{8}} \)
(b)
Method:
\( G(1) = \frac{1}{16}(1) + \frac{3}{16}(1)^2 + \frac{5}{16}(1)^3 + \frac{7}{16}(1)^4 \)
\( = \frac{1+3+5+7}{16} = \frac{16}{16} = 1 \)
\( \boxed{1} \)
(c) (i)
Method:
\( G'(t) = \frac{d}{dt} \left( \frac{1}{16}t + \frac{3}{16}t^2 + \frac{5}{16}t^3 + \frac{7}{16}t^4 \right) \)
\( = \frac{1}{16} + \frac{6}{16}t + \frac{15}{16}t^2 + \frac{28}{16}t^3 \)
\( = \frac{1}{16} + \frac{3}{8}t + \frac{15}{16}t^2 + \frac{7}{4}t^3 \)
\( \boxed{G'(t) = \frac{1}{16} + \frac{3}{8}t + \frac{15}{16}t^2 + \frac{7}{4}t^3} \)
(c) (ii)
Method:
Substitute \( t = 1 \) into \( G'(t) \):
\( G'(1) = \frac{1}{16} + \frac{3}{8} + \frac{15}{16} + \frac{7}{4} \)
\( = \frac{1}{16} + \frac{6}{16} + \frac{15}{16} + \frac{28}{16} = \frac{50}{16} = \frac{25}{8} \)
From part (a), \( E(M) = \frac{25}{8} \), so \( G'(1) = E(M) \).
\( \boxed{G'(1) = E(M)} \)
(d)
Method 1 (Direct probabilities):
\( P(X=0) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10} \)
\( P(X=1) = \frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5} \)
\( P(X=2) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10} \)
Alternatively: \( P(X=2) = 1 – \left( \frac{3}{10} + \frac{3}{5} \right) = 1 – \frac{9}{10} = \frac{1}{10} \)
Thus \( G_X(t) = \frac{3}{10} + \frac{3}{5}t + \frac{1}{10}t^2 \).
\( \boxed{G_X(t) = \frac{3}{10} + \frac{3}{5}t + \frac{1}{10}t^2} \)
(e) (i)
Method:
\( P(Y=2) = P(\text{tail on biased}) \times P(\text{tail on unbiased}) = p \times \frac{1}{2} = \frac{p}{2} \)
Given \( \frac{p}{2} = \frac{1}{3} \) ⇒ \( p = \frac{2}{3} \).
\( \boxed{\frac{2}{3}} \)
(e) (ii)
Method:
\( P(Y=0) = P(\text{head on both}) = (1-p) \times \frac{1}{2} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \)
\( P(Y=1) = P(\text{tail on biased only}) + P(\text{tail on unbiased only}) = \frac{2}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{2} = \frac{1}{2} \)
Thus \( G_Y(t) = \frac{1}{6} + \frac{1}{2}t + \frac{1}{3}t^2 \).
\( \boxed{G_Y(t) = \frac{1}{6} + \frac{1}{2}t + \frac{1}{3}t^2} \)
(f)
Method 1 (Differentiate product and evaluate at \( t=1 \)):
\( G_z(t) = \left( \frac{3}{10} + \frac{3}{5}t + \frac{1}{10}t^2 \right) \left( \frac{1}{6} + \frac{1}{2}t + \frac{1}{3}t^2 \right) \)
Expand: \( G_z(t) = \frac{1}{20} + \frac{1}{4}t + \frac{5}{12}t^2 + \frac{1}{4}t^3 + \frac{1}{30}t^4 \)
Differentiate: \( G_z'(t) = \frac{1}{4} + \frac{5}{6}t + \frac{3}{4}t^2 + \frac{2}{15}t^3 \)
Evaluate at \( t=1 \): \( G_z'(1) = \frac{1}{4} + \frac{5}{6} + \frac{3}{4} + \frac{2}{15} = \frac{59}{30} \approx 1.9666… \)
Thus \( E(Z) = \frac{59}{30} \).
Method 2 (Use derivative of product rule directly):
\( G_z'(t) = G_X'(t)G_Y(t) + G_X(t)G_Y'(t) \)
\( G_X'(t) = \frac{3}{5} + \frac{1}{5}t \), \( G_Y'(t) = \frac{1}{2} + \frac{2}{3}t \)
Then substitute \( t=1 \) to find \( G_z'(1) \).
Method 3 (Use additive property of expectation):
Since \( Z = X + Y \) and \( G_z(t) = G_X(t)G_Y(t) \), we have \( E(Z) = E(X) + E(Y) \).
\( E(X) = G_X'(1) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} \)
\( E(Y) = G_Y'(1) = \frac{1}{2} + \frac{2}{3} = \frac{7}{6} \)
\( E(Z) = \frac{4}{5} + \frac{7}{6} = \frac{24}{30} + \frac{35}{30} = \frac{59}{30} \)
\( \boxed{\frac{59}{30}} \)