IB Mathematics SL 4.5 Concepts of trial, outcome AA SL Paper 1- Exam Style Questions- New Syllabus

(a) Show \( k = \frac{1}{3} \):
The sum of all probabilities in a discrete distribution must equal 1:
\( k + 3k^2 + 2k^2 + k^2 = 1 \)
\( 6k^2 + k – 1 = 0 \)
Solve the quadratic equation:
\( (3k – 1)(2k + 1) = 0 \)
\( k = \frac{1}{3} \) or \( k = -\frac{1}{2} \).
Since probabilities must be non-negative and \( k > 0 \), \( \boxed{k = \frac{1}{3}} \). (Shown)

(b) Find \( P(X < 3a) \):
\( P(X < 3a) = 1 – P(X = 3a) \)
\( = 1 – k^2 = 1 – \left( \frac{1}{3} \right)^2 = 1 – \frac{1}{9} \)
Answer: \( \boxed{\frac{8}{9}} \)

(c) Find the conditional probability:
\( P(X \geq a \mid X < 3a) = \frac{P(a \leq X < 3a)}{P(X < 3a)} \)
The numerator is \( P(X=a) + P(X=2a) = 3k^2 + 2k^2 = 5k^2 = \frac{5}{9} \).
From part (b), the denominator is \( \frac{8}{9} \).
\( \frac{5/9}{8/9} = \frac{5}{8} \)
Answer: \( \boxed{0.625 \text{ (or } \frac{5}{8})} \)

(d) Find the value of \( a \):
\( E(X) = \sum x \cdot P(X = x) \)
\( E(X) = (0)(k) + (a)(3k^2) + (2a)(2k^2) + (3a)(k^2) \)
\( E(X) = 3ak^2 + 4ak^2 + 3ak^2 = 10ak^2 \)
Substitute \( k^2 = \frac{1}{9} \):
\( \frac{10a}{9} = 20 \implies 10a = 180 \implies a = 18 \)
Answer: \( \boxed{18} \)