IB Mathematics SL 4.7 Concept of discrete random variables AI SL Paper 1- Exam Style Questions- New Syllabus
Question
(i) \( P(X = 0) \).
(ii) \( P(X = 5) \).
Most appropriate topic codes (IB Mathematics: applications and interpretation):
• SL 4.7: Discrete random variables and their probability distributions; expected value \( E(X) \) — part (c)
▶️ Answer/Explanation
(a)
(i) \( P(X=0) = \frac{6}{36} = \frac{1}{6} \) (outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6))
(ii) \( P(X=5) = \frac{2}{36} = \frac{1}{18} \) (outcomes: (1,6), (6,1))
✅ Answer: (i) \( \boxed{\frac{1}{6}} \), (ii) \( \boxed{\frac{1}{18}} \)
(b)
Complete distribution based on the 36 possible outcomes:
\( P(0)=\frac{6}{36},\; P(1)=\frac{10}{36},\; P(2)=\frac{8}{36},\; P(3)=\frac{6}{36},\; P(4)=\frac{4}{36},\; P(5)=\frac{2}{36} \)
Table:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(X = x) & \frac{6}{36} & \frac{10}{36} & \frac{8}{36} & \frac{6}{36} & \frac{4}{36} & \frac{2}{36} \\ \hline \end{array} \]
(c)
\( E(X) = \sum x \cdot P(X=x) \)
\( E(X) = 0\left(\frac{6}{36}\right)+1\left(\frac{10}{36}\right)+2\left(\frac{8}{36}\right)+3\left(\frac{6}{36}\right)+4\left(\frac{4}{36}\right)+5\left(\frac{2}{36}\right) \)
\( E(X) = \frac{0+10+16+18+16+10}{36} = \frac{70}{36} = \frac{35}{18} \)
✅ Answer: \( \boxed{\frac{35}{18}} \)
(d)
Odd differences are \( X \in \{1, 3, 5\} \).
Total probability of an odd difference: \( \frac{10}{36} + \frac{6}{36} + \frac{2}{36} = \frac{18}{36} \).
Conditional probability: \( P(X=5 \mid X \text{ is odd}) = \frac{P(X=5)}{P(X \in \{1,3,5\})} = \frac{2/36}{18/36} = \frac{2}{18} = \frac{1}{9} \).
✅ Answer: \( \boxed{\frac{1}{9}} \)