IB Mathematics SL 4.11 Formal definition and use of the formulae AA SL Paper 1- Exam Style Questions- New Syllabus

Given information:
\( P(A’) = \frac{3}{4} \), \( P(A \cup B) = \frac{3}{4} \), \( P(B|A) = \frac{2}{3} \)

(a) Finding \( P(A \cap B) \)
Step 1: Find \( P(A) \) from \( P(A’) \)
\( P(A) = 1 – P(A’) = 1 – \frac{3}{4} = \frac{1}{4} \)

Step 2: Use conditional probability formula:
\( P(B|A) = \frac{P(A \cap B)}{P(A)} \)
\( \frac{2}{3} = \frac{P(A \cap B)}{\frac{1}{4}} \)

Step 3: Solve for \( P(A \cap B) \):
\( P(A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6} \)

\( \boxed{P(A \cap B) = \dfrac{1}{6}} \)

(b) Showing \( A \) and \( B \) are independent
Step 1: Find \( P(B) \) using addition rule:
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)
\( \frac{3}{4} = \frac{1}{4} + P(B) – \frac{1}{6} \)

Step 2: Solve for \( P(B) \):
\( P(B) = \frac{3}{4} – \frac{1}{4} + \frac{1}{6} \)
\( P(B) = \frac{2}{4} + \frac{1}{6} = \frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \)

Step 3: Check independence condition:
Two equivalent methods:

Method 1: Compare \( P(B|A) \) with \( P(B) \):
\( P(B|A) = \frac{2}{3} \) (given)
\( P(B) = \frac{2}{3} \) (calculated)
Since \( P(B|A) = P(B) \), events are independent.

Method 2: Check \( P(A \cap B) = P(A)P(B) \):
\( P(A)P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6} \)
\( P(A \cap B) = \frac{1}{6} \) (from part a)
Since \( P(A \cap B) = P(A)P(B) \), events are independent.

Events \( A \) and \( B \) are independent as shown above.