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IB Mathematics AHL 4.13 Use of Bayes’ theorem for a maximum of three events AA HL Paper 3 | Exam Style Question

IB Mathematics AHL 4.13 Use of Bayes’ theorem for a maximum of three events AA HL Paper 3

IB DP Maths AA: HL Papers 3: All Topics
Question

This question considers two models for random boosts in a computer game, where each action has a chance of being boosted, providing a benefit to the player.

In the first model, the probability of a boost is constant. In the second model, the boost probability increases by 0.2 after each non-boosted action, resetting to 0.2 after a boost.

(a) Suppose the probability of a boost is 0.1.

(i) Find the probability that the first boost occurs on the third action [2].

(ii) Find the probability that at least one boost occurs in the first six actions [3].

(b) Suppose the probability of a boost is \( p \), where \( 0 < p < 1 \). Let \( X \) be the number of actions until the first boost.

(i) Explain why the probability that the first boost occurs on the \( x \)-th action is \( p(1-p)^{x-1} \) [2].

(ii) Write an expression, using sigma notation, for \( E(X) \) in terms of \( x \) and \( p \) [2].

(c) Consider the infinite geometric series with first term \( a \) and common ratio \( r \), \( |r| < 1 \):

\[ a + ar + ar^2 + ar^3 + \dots = \frac{a}{1-r} \]

(i) By differentiating both sides with respect to \( r \), find an expression for \( \sum_{n=1}^{\infty} n a r^{n-1} \) [3].

(ii) Hence, show that \( E(X) = \frac{1}{p} \) [2].

(d) Given \( \text{Var}(X) = \frac{1-p}{p^2} \), find \( E(X) \) and \( \text{Var}(X) \) when \( p = 0.1 \) [2].

(e) In the second model, the initial boost probability is 0.2, increasing by 0.2 per non-boosted action, resetting to 0.2 after a boost. Show that the probability of the first boost on the third action is 0.288 [3].

(f) Let \( Y \) be the number of actions until the first boost in the second model. Explain why \( Y \leq 5 \) [2].

(g) For the second model, the probability distribution of \( Y \) is:

\[ \begin{array}{|c|c|c|c|c|c|} \hline y & 1 & 2 & 3 & 4 & 5 \\ \hline P(Y = y) & 0.2 & m & 0.288 & n & 0.0384 \\ \hline \end{array} \]

(i) Find the values of \( m \) and \( n \) [2].

(ii) Show that \( E(Y) = 2.5104 \) [2].

(iii) Find \( \text{Var}(Y) \) [2].

(h) For the two models:

(i) Use the expression from (c)(ii) to find \( p \) such that \( E(X) = E(Y) \) [2].

(ii) Find \( \text{Var}(X) \) for this \( p \) [2].

(iii) Determine, with a reason, which model provides a more consistent experience for boosted actions [2].

▶️ Answer/Explanation
Markscheme Solution

Note: Model 1 uses a geometric distribution with constant boost probability \( p \). Model 2 has boost probability starting at 0.2, increasing by 0.2 per non-boosted action, resetting to 0.2 after a boost. Numerical answers are rounded to 3 significant figures unless specified.

(a)(i)

First boost on third action (\( p = 0.1 \)):

No boost on first two actions (\( 1 – 0.1 = 0.9 \)), boost on third (0.1) (M1).

\[ P = 0.9^2 \times 0.1 = 0.81 \times 0.1 = 0.081 \] (A1)

[2 marks]

(a)(ii)

At least one boost in six actions:

Use complement: \( P(\text{at least one}) = 1 – P(\text{no boosts}) \) (M1).

\[ P(\text{no boost}) = 0.9 \implies P(\text{no boosts in 6}) = 0.9^6 = 0.531441 \] (A1).

\[ P(\text{at least one}) = 1 – 0.531441 = 0.468559 \approx 0.469 \] (A1).

[3 marks]

(b)(i)

Probability of first boost on \( x \)-th action:

\( x-1 \) failures (\( 1-p \)), one success (\( p \)) (M1).

\[ P(X = x) = (1-p)^{x-1} \times p \] (A1).

[2 marks]

(b)(ii)

Expected value \( E(X) \):

\[ E(X) = \sum_{x=1}^\infty x \cdot P(X = x) = \sum_{x=1}^\infty x p (1-p)^{x-1} \] (M1)(A1).

[2 marks]

(c)(i)

Given: \( \sum_{n=0}^\infty ar^n = \frac{a}{1-r} \), \( |r| < 1 \).

Differentiate: Left: \( \sum_{n=1}^\infty n a r^{n-1} \); Right: \( \frac{d}{dr} \left( \frac{a}{1-r} \right) = \frac{a}{(1-r)^2} \) (M1)(A1).

\[ \sum_{n=1}^\infty n a r^{n-1} = \frac{a}{(1-r)^2} \] (A1).

[3 marks]

(c)(ii)

For \( E(X) = \sum_{x=1}^\infty x p (1-p)^{x-1} \), set \( a = p \), \( r = 1-p \):

\[ p \sum_{x=1}^\infty x (1-p)^{x-1} = p \cdot \frac{1}{(1 – (1-p))^2} = p \cdot \frac{1}{p^2} = \frac{1}{p} \] (M1)(A1).

[2 marks]

(d)

For \( p = 0.1 \):

\[ E(X) = \frac{1}{0.1} = 10 \] (A1).

\[ \text{Var}(X) = \frac{1 – 0.1}{(0.1)^2} = \frac{0.9}{0.01} = 90 \] (A1).

[2 marks]

(e)

Second model, first boost on third action:

Action 1: no boost (\( 0.8 \)); Action 2: no boost (\( 0.6 \)); Action 3: boost (\( 0.6 \)) (M1).

\[ P(Y = 3) = 0.8 \times 0.6 \times 0.6 = 0.288 \] (A1)(A1).

[3 marks]

(f)

Boost probabilities: 0.2, 0.4, 0.6, 0.8, 1.0 for actions 1–5 (M1).

At action 5, \( P(\text{boost}) = 1.0 \), so \( Y \leq 5 \) (A1).

[2 marks]

(g)(i)

Probability distribution: \( P(Y = 1) = 0.2 \), \( P(Y = 3) = 0.288 \), \( P(Y = 5) = 0.0384 \).

\( P(Y = 2) = 0.8 \times 0.4 = 0.32 = m \).

\[ P(Y = 4) = 0.8 \times 0.6 \times 0.4 \times 0.8 = 0.1536 = n \] (M1).

\[ m = 0.32, n = 0.1536 \] (A1).

[2 marks]

(g)(ii)

\[ E(Y) = 1 \cdot 0.2 + 2 \cdot 0.32 + 3 \cdot 0.288 + 4 \cdot 0.1536 + 5 \cdot 0.0384 \]

\[ = 0.2 + 0.64 + 0.864 + 0.6144 + 0.192 = 2.5104 \] (M1)(A1).

[2 marks]

(g)(iii)

\[ E(Y^2) = 1^2 \cdot 0.2 + 2^2 \cdot 0.32 + 3^2 \cdot 0.288 + 4^2 \cdot 0.1536 + 5^2 \cdot 0.0384 \]

\[ = 0.2 + 1.28 + 2.592 + 2.4576 + 0.96 = 7.4896 \] (M1).

\[ \text{Var}(Y) = 7.4896 – (2.5104)^2 = 7.4896 – 6.3026816 = 1.1869184 \approx 1.1869 \] (A1).

[2 marks]

(h)(i)

\[ E(X) = \frac{1}{p} = E(Y) = 2.5104 \implies p = \frac{1}{2.5104} \approx 0.3982 \approx 0.398 \] (M1)(A1).

[2 marks]

(h)(ii)

For \( p = 0.3982 \):

\[ \text{Var}(X) = \frac{1 – 0.3982}{(0.3982)^2} = \frac{0.6018}{0.1586} \approx 3.7933 \approx 3.79 \] (M1)(A1).

[2 marks]

(h)(iii)

Compare: \(\text{Var}(X) \approx 3.79\), \(\text{Var}(Y) \approx 1.1869\). Lower variance means more consistent boost timing (M1).

Second model is more consistent (A1).

[2 marks]

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