IBDP Maths SL 4.5 Concepts of trial, outcome AA HL Paper 1- Exam Style Questions- New Syllabus

Question

Two events \( A \) and \( B \) satisfy the conditions \( P(A’) = P(A \cup B) = \frac{3}{4} \) and \( P(B \mid A) = \frac{2}{3} \).

(a) Determine the value of \( P(A \cap B) \).

(b) Hence, show that the events \( A \) and \( B \) are independent.

Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):

• SL 4.5: Conditional probability, including the use of \( P(A \mid B) \) — parts (a), (b)
• SL 4.6: Independent events; use of \( P(A \cap B) = P(A)P(B) \) — part (b)

▶️ Answer/Explanation

(a) Finding \( P(A \cap B) \):

Since \( P(A’) = \frac{3}{4} \), it follows that \( P(A) = 1 – \frac{3}{4} = \frac{1}{4} \).
Using the conditional probability formula, \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \).
Substituting the given values, \( \frac{2}{3} = \frac{P(A \cap B)}{\frac{1}{4}} \).
Hence, \( P(A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} \).
\( \boxed{P(A \cap B) = \frac{1}{6}} \)

(b) Showing that \( A \) and \( B \) are independent:

First find \( P(B) \) using the addition rule:
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \).
Substituting known values, \( \frac{3}{4} = \frac{1}{4} + P(B) – \frac{1}{6} \).
Solving, \( P(B) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} \).

Now check independence:
\( P(A)P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} \).
Since this equals \( P(A \cap B) \), the events \( A \) and \( B \) are independent.
\( \boxed{\text{Independent}} \)

Question

Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.4\).

a. Find the value of \({\text{P}}(A \cup B)\) when

(i) \(A\) and \(B\) are mutually exclusive;

(ii) \(A\) and \(B\) are independent. [4]

b. Given that \({\text{P}}(A \cup B) = 0.6\), find \({\text{P}}(A|B)\). [3]

▶️ Answer/Explanation

Solution a(i)

\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1

Solution a(ii)

\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) M1

\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) M1

\( = 0.3 + 0.4 – 0.12 = 0.58\) A1

[4 marks]

Solution b

\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cup B)\)

\( = 0.3 + 0.4 – 0.6 = 0.1\) A1

\({\text{P}}(A|B) = \frac{{\text{P}}(A \cap B)}{{\text{P}}(B)}\) M1

\( = \frac{0.1}{0.4} = 0.25\) A1

[3 marks]

IBDP Maths SL 4.5 Concepts of trial, outcome AA HL Paper 1- Exam Style Questions- New Syllabus

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