Mathematics · Pearson Edexcel 9MA0 · Topic 3

Probability

Probability with Venn diagrams, tree diagrams and conditional probability.

  • 3 sub-topics
  • 1 in Year 12, 2 in Year 13
  • 41 practice questions
  • 3 animated lessons
Practice questions (41) Lessons Worksheets All Edexcel Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.1Mutually exclusive and independent eventsY12
  2. 3.2Conditional probabilityY13
  3. 3.3Modelling with probabilityY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 41 questions in this topic, free, no account needed.

Probability

Easy

Exclusive or independent? · Year 12 · 3 marks

P(A)=0.3\mathrm{P}(A) = 0.3 and P(B)=0.5\mathrm{P}(B) = 0.5
  1. (a)
    Find P(A∪B)\mathrm{P}(A \cup B) if AA and BB are mutually exclusive.
    (1)
  2. (b)
    Find P(A∪B)\mathrm{P}(A \cup B) if instead AA and BB are independent.
    (2)
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Part (a)1 mark

  1. Mutually exclusive events cannot both happen, so just add.0.3+0.5=0.80.3 + 0.5 = 0.8

Part (b)2 marks

  1. Independent: multiply.P(A∩B)\mathrm{P}(A \cap B)=0.3×0.5{}= 0.3 \times 0.5=0.15{}= 0.15
  2. Take away the overlap so it is not counted twice.P(A∪B)\mathrm{P}(A \cup B)=0.3+0.5−0.15{}= 0.3 + 0.5 - 0.15=0.65{}= 0.65

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Probability

Not needed for Year 12Medium

Conditional probability · Year 13 · 3 marks

P(A)=0.4\mathrm{P}(A) = 0.4, P(B)=0.5\mathrm{P}(B) = 0.5 and P(A∩B)=0.15\mathrm{P}(A \cap B) = 0.15.
  1. Find P(A∣B)\mathrm{P}(A \mid B) and P(B∣A)\mathrm{P}(B \mid A).
    (3)
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  1. "Given BB" means we only look inside BB.P(A∣B)\mathrm{P}(A \mid B)=P(A∩B)P(B){}= \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}
  2. =0.150.5=0.3= \dfrac{0.15}{0.5} = 0.3
  3. P(B∣A)\mathrm{P}(B \mid A)=0.150.4{}= \dfrac{0.15}{0.4}=0.375{}= 0.375

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Probability

Not needed for Year 12Hard

Reversing a condition · Year 13 · 5 marks

P(A)=0.5\mathrm{P}(A) = 0.5, P(B∣A)=0.4\mathrm{P}(B \mid A) = 0.4 and P(B∣A′)=0.2\mathrm{P}(B \mid A') = 0.2.
  1. (a)
    Find P(B)\mathrm{P}(B).
    (3)
  2. (b)
    Find P(A∣B)\mathrm{P}(A \mid B).
    (2)

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Part (a)3 marks

  1. P(A∩B)\mathrm{P}(A \cap B)=0.5×0.4{}= 0.5 \times 0.4=0.2{}= 0.2
  2. P(A′∩B)\mathrm{P}(A' \cap B)=0.5×0.2{}= 0.5 \times 0.2=0.1{}= 0.1
  3. P(B)=0.3\mathrm{P}(B) = 0.3

Part (b)2 marks

  1. 0.20.3\dfrac{0.2}{0.3}
  2. =23= \dfrac{2}{3}

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