17N.2.AHL.TZ0.H_2

pestleMathematicsAIHLPaper 217N· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

Events $A$ and $B$ are such that ${\text{P}}(A \cup B) = 0.95,{\text{ P}}(A \cap B) = 0.6$ and ${\text{P}}(A|B) = 0.75$ .

Find ${\text{P}}(B)$ .

[2]

Find ${\text{P}}(A)$ .

[2]

Hence show that events $A^{'}$ and $B$ are independent.

[2]

Markscheme / solution

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

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

$\Rightarrow 0.75 = \frac{{0.6}}{{{\text{P}}(B)}}$ (M1)

$\Rightarrow {\text{P}}(B){\text{ }}\left( { = \frac{{0.6}}{{0.75}}} \right) = 0.8$ A1

[2 marks]

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

$\Rightarrow 0.95 = {\text{P}}(A) + 0.8 - 0.6$ (M1)

$\Rightarrow {\text{P}}(A) = 0.75$ A1

[2 marks]

METHOD 1

${\text{P}}(A'|B) = \frac{{{\text{P}}(A' \cap B)}}{{{\text{P}}(B)}} = \frac{{0.2}}{{0.8}} = 0.25$ A1

${\text{P}}(A'|B) = {\text{P}}(A’)$ R1

hence $A^{'}$ and $B$ are independent AG

Note: If there is evidence that the student has calculated ${\text{P}}(A' \cap B) = 0.2$ by assuming independence in the first place, award A0R0.

METHOD 2

EITHER

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

${\text{P}}(A) \times {\text{P}}(B) = 0.75 \times 0.80 = 0.6 = {\text{P}}(A \cap B)$ A1

THEN

$A$ and $B$ are independent R1

hence $A^{'}$ and $B$ are independent AG

METHOD 3

${\text{P}}(A') \times {\text{P}}(B) = 0.25 \times 0.80 = 0.2$ A1

${\text{P}}(A') \times {\text{P}}(B) = {\text{P}}(A' \cap B)$ R1

hence $A^{'}$ and $B$ are independent AG

[2 marks]

Examiners’ report

[N/A]