IBDP Maths SL 4.5 Concepts of trial, outcome AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) 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]
\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cup B)\)
\( = 0.3 + 0.4 – 0.6 = 0.1\) 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]
Question
Events \( A \) and \( B \) are such that
\(
P(A) = 0.65, \quad P(B) = 0.75, \quad \text{and} \quad P(A \cap B) = 0.6.
\)
(a) Find \( P(A \cup B) \).
(b) Hence, or otherwise, find \( P(A’ \cap B’) \).
▶️Answer/Explanation
(a) Finding \( P(A \cup B) \):
Using the formula for the union of two events:
\(
P(A \cup B) = P(A) + P(B) – P(A \cap B)
\)
Substituting the given values:
\(
P(A \cup B) = 0.65 + 0.75 – 0.6
\)
\(
P(A \cup B) = 0.80
\)
(b) Finding \( P(A’ \cap B’) \):
Using the complement rule:
\(
P(A’ \cap B’) = 1 – P(A \cup B)
\)
Substituting \( P(A \cup B) = 0.80 \):
\(
P(A’ \cap B’) = 1 – 0.80
\)
\(
P(A’ \cap B’) = 0.20
\)
Thus, the required probability is \( P(A’ \cap B’) = 0.20 \).