Home / IB Mathematics SL 4.8 Binomial distribution, its mean and variance AA SL Paper 2- Exam Style Questions

IB Mathematics SL 4.8 Binomial distribution, its mean and variance AA SL Paper 2- Exam Style Questions- New Syllabus

IB DP Maths AA SL Paper 2 - All Topics

Question

Lynn participates in a game using a pair of fair six-sided dice, each labeled 1 through 6. For every round played, she tosses both dice once. Points are allocated based on the resulting values:

Two dice showing different values

 
  • 10 points if both dice display the same number.
  • 5 points if the numerical difference between the two dice is exactly 1 (e.g., 2 and 3).
  • 0 points for any other combination.
Let the discrete random variable \( X \) represent the points Lynn earns in a single round.
(a) Show that the probability of Lynn scoring 5 points in a round is \( \frac{5}{18} \).
(b) Calculate the probability of scoring 0 points in a single round.
(c) Find the expected value \( E(X) \).
(d) Use your answer from part (c) to estimate Lynn’s total score after 90 rounds.
(e) Lynn plays exactly five rounds. A prize is given if the total score exceeds 40 points. Find the probability that Lynn wins the prize.

Most-appropriate topic codes (Mathematics: analysis and approaches guide):

• SL 4.5: Concepts of trial, outcome, and sample space; \( P(A) = \frac{n(A)}{n(U)} \) — Part a, b
• SL 4.6: Use of tables of outcomes to calculate probabilities — Part b, e
• SL 4.7: Discrete random variables and their probability distributions; expected value \( E(X) \) — Part c
• SL 4.8: Situations where the binomial distribution is an appropriate model — Part d, e
▶️ Answer/Explanation

(a)
Total possible outcomes when throwing two dice: \( 6 \times 6 = 36 \).
Outcomes where the difference is 1:
(1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5).
There are 10 such outcomes.
\( P(X=5) = \frac{10}{36} = \frac{5}{18} \).

(b)
Identify outcomes for other point values:
– 10 points (same score): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes.
– 5 points (difference of 1): 10 outcomes (found in part a).
Remaining outcomes for 0 points: \( 36 – (6 + 10) = 20 \).
\( P(X=0) = \frac{20}{36} = \frac{5}{9} \).

(c)
The probability distribution for \( X \) is:
\( P(X=0) = \frac{20}{36} \), \( P(X=5) = \frac{10}{36} \), \( P(X=10) = \frac{6}{36} \).
\( E(X) = \sum x \cdot P(X=x) \)
\( E(X) = \left( 0 \cdot \frac{20}{36} \right) + \left( 5 \cdot \frac{10}{36} \right) + \left( 10 \cdot \frac{6}{36} \right) \)
\( E(X) = 0 + \frac{50}{36} + \frac{60}{36} = \frac{110}{36} = \frac{55}{18} \approx 3.06 \).

(d)
Expected total score = Number of rounds \(\times E(X)\).
Total points \( = 90 \times \frac{55}{18} = 5 \times 55 = 275 \).

(e)
To win the prize (total > 40 points) in 5 rounds, Lynn must score either:
1. 50 points: (10, 10, 10, 10, 10) → all five rounds are 10-point rounds.
2. 45 points: One 5-point round and four 10-point rounds.

Let \( p_{10} = \frac{6}{36} = \frac{1}{6} \) and \( p_{5} = \frac{10}{36} = \frac{5}{18} \).
– Prob(50 pts) \( = (p_{10})^5 = \left( \frac{1}{6} \right)^5 = \frac{1}{7776} \).
– Prob(45 pts) \( = \binom{5}{1} \times (p_5)^1 \times (p_{10})^4 = 5 \times \frac{5}{18} \times \left( \frac{1}{6} \right)^4 = \frac{25}{23328} \).
Total probability \( = \frac{3}{23328} + \frac{25}{23328} = \frac{28}{23328} = \frac{7}{5832} \approx 0.00120 \).

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