16N.1.AHL.TZ0.H_10

pestleMathematicsAAHLPaper 116N· sl-1-3-geometric-sequences-and-series, sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

Consider two events $A$ and $A$ defined in the same sample space.

Given that ${\text{P}}(A \cup B) = \frac{4}{9},{\text{ P}}(B|A) = \frac{1}{3}$ and ${\text{P}}(B|A') = \frac{1}{6}$ ,

Show that ${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)$ .

[3]

(i) show that ${\text{P}}(A) = \frac{1}{3}$ ;

(ii) hence find ${\text{P}}(B)$ .

[6]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

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

$= {\text{P}}(A) + {\text{P}}(A \cap B) + {\text{P}}(A' \cap B) - {\text{P}}(A \cap B)$ M1A1

$= {\text{P}}(A) + {\text{P}}(A' \cap B)$ AG

METHOD 2

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

$= {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A|B) \times {\text{P}}(B)$ M1

$= {\text{P}}(A) + \left( {1 - {\text{P}}(A|B)} \right) \times {\text{P}}(B)$

$= {\text{P}}(A) + {\text{P}}(A'|B) \times {\text{P}}(B)$ A1

$= {\text{P}}(A) + {\text{P}}(A' \cap B)$ AG

[3 marks]

(i) use ${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)$ and ${\text{P}}(A' \cap B) = {\text{P}}(B|A'){\text{P}}(A')$ (M1)

$\frac{4}{9} = {\text{P}}(A) + \frac{1}{6}\left( {1 - {\text{P}}(A)} \right)$ A1

$8 = 18{\text{P}}(A) + 3\left( {1 - {\text{P}}(A)} \right)$ M1

${\text{P}}(A) = \frac{1}{3}$ AG

(ii) METHOD 1

${\text{P}}(B) = {\text{P}}(A \cap B) + {\text{P}}(A' \cap B)$ M1

$= {\text{P}}(B|A){\text{P}}(A) + {\text{P}}(B|A'){\text{P}}(A')$ M1

$= \frac{1}{3} \times \frac{1}{3} + \frac{1}{6} \times \frac{2}{3} = \frac{2}{9}$ A1

METHOD 2

${\text{P}}(A \cap B) = {\text{P}}(B|A){\text{P}}(A) \Rightarrow {\text{P}}(A \cap B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ M1

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

${\text{P}}(B) = \frac{4}{9} + \frac{1}{9} - \frac{1}{3} = \frac{2}{9}$ A1

[6 marks]

Examiners’ report

[N/A]