OCR MEI Paper 2 Specimen — Question 7 4 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks4
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Mark schemeDownload PDF ↗
TopicConditional Probability
TypeIndependence test with conditional probability
DifficultyModerate -0.8 This is a straightforward probability question requiring only the addition rule P(A∪B) = P(A) + P(B) - P(A∩B) to find P(A∩B) = 0.25, then applying the conditional probability formula P(A|B) = P(A∩B)/P(B) = 0.5. It's a direct two-step application of standard formulas with no problem-solving insight required, making it easier than average.
Spec2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles
Two events \(A\) and \(B\) are such that \(\text{P}(A) = 0.6\), \(\text{P}(B) = 0.5\) and \(\text{P}(A \cup B) = 0.85\). Find \(\text{P}(A | B)\). [4]
Question 7:
AnswerMarks
7P(A∩B) = P(A) + P(B) – P(A∪B)
= 0.6 + 0.5 – 0.85
= 0.25
P(A(cid:136)B)
AnswerMarks
P(cid:11)AB(cid:12) (cid:32)
P(cid:11)B(cid:12)
0.25
(cid:32)
0.5
AnswerMarks
= 0.5M1
A1
M1
A1
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
1.1n
e
AnswerMarks Guidance
73 0 
Question 7:
7 | P(A∩B) = P(A) + P(B) – P(A∪B)
= 0.6 + 0.5 – 0.85
= 0.25
P(A(cid:136)B)
P(cid:11)A|B(cid:12) (cid:32)
P(cid:11)B(cid:12)
0.25
(cid:32)
0.5
= 0.5 | M1
A1
M1
A1
[4] | 3.1a
1.1
1.1
1.1 | n
e
7 | 3 | 0 | 1 | 0 | 4 | 0
Two events $A$ and $B$ are such that $\text{P}(A) = 0.6$, $\text{P}(B) = 0.5$ and $\text{P}(A \cup B) = 0.85$. Find $\text{P}(A | B)$. [4]

\hfill \mbox{\textit{OCR MEI Paper 2  Q7 [4]}}
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Q1 5 Q2 4 Q3 3 Q4 5 Q5 2 Q6 4 Q7 4 Q8 3 Q9 4 Q10 3 Q11 4 Q12 6 Q13 6 Q14 12 Q15 15 Q16 20