Conditional probability: 7 exam questions solved step by step

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What this lesson covers

  • A conditional probability restricts attention to one group, so its denominator is that group's total.
  • The notation P(A | B) is read 'A given B', and the order matters.
Key idea

7 questions, one idea

  • 1. "Given B": B is the whole world.
  • 2. Divide by the total of B.
  • 3. P(A | B) = P(A and B) ÷ P(B)
  • 4. Notation, 4 d.p., then context.
Your turn

Question 1 · Achieved

  • 320 visitors to a museum.
  • Find P(ate at the café | adult).
CaféNo caféTotal
Adult72108180
Child28112140
Total100220320
Worked example

Q1: P(C | A)

  • Given adult: adults only.
  • Divide by the row total.
  1. A: adult, C: ate at the café
  2. P(C | A)given adult
  3. adults: 180the reduced total
  4. P(C | A) = 72 ÷ 180
  5. = 0.4
  6. 72 ÷ 320 = 0.225that is P(A and C)
  7. 40% of adults ate at the café
Your turn

Question 2 · Achieved

  • Same museum table.
  • Find P(adult | ate at the café).
  • Careful: the order has changed.
CaféNo caféTotal
Adult72108180
Child28112140
Total100220320
Worked example

Q2: P(A | C)

  • Given café: café column.
  • Order matters.
  1. P(A | C)given café
  2. café visitors: 100the reduced total
  3. P(A | C) = 72 ÷ 100 = 0.72
  4. P(C | A) = 0.4question 1
  5. P(A | C) ≠ P(C | A)order matters
  6. P(A and C) = 72 ÷ 320 = 0.225
  7. 72% of café visitors: adults
Your turn

Question 3 · Merit

  • Customers of an online store:
  • bought a book (B), a game (G).
  • Find P(B | G) and P(G | not B).
Question 3 · Merit
Worked example

Q3: the given region

  • Given G: the G circle.
  • Not B: outside B.
  1. P(G) = 0.09 + 0.21 = 0.3the games circle
  2. P(B | G) = 0.09 ÷ 0.3both ÷ P(G)
  3. = 0.3
  4. P(not B) = 0.21 + 0.37outside books
  5. = 0.58
  6. P(G | not B) = 0.21 ÷ 0.58only G ÷ P(not B)
  7. = 0.36214 d.p.
Your turn

Question 4 · Merit

  • On school days in a town:
  • P(rain) = 0.35, P(bus late) = 0.5,
  • P(rain or bus late) = 0.65.
  • Find P(bus late | rain).
Worked example

Q4: P(L | R) from "or"

  • Find "and" first.
  • Then divide by P(R).
  1. P(R or L) = P(R) + P(L)addition rule
  2. − P(R and L)
  3. 0.65 = 0.35 + 0.5 − P(R and L)
  4. P(R and L) = 0.85 − 0.65 = 0.2
  5. P(L | R) = P(R and L) ÷ P(R)
  6. = 0.2 ÷ 0.35
  7. = 0.57144 d.p.
  8. P(R | L) = 0.2 ÷ 0.5 = 0.4wrong way round
Your turn

Question 5 · Merit, in context

  • At an airport, 4% of bags hold food.
  • A dog flags 90% of bags with food
  • and 6% of bags without.
  • Find P(a bag is flagged).
Worked example

Q5: build the tree

  • First: food 0.04, none 0.96.
  • Second branches are conditional.
  • Each pair adds to 1.
Q5: build the tree
Worked example

Q5: P(flag)

  • Two paths end in a flag.
  • Multiply, then add.
  1. food, flag: 0.04 × 0.9 = 0.036
  2. no food, flag: 0.96 × 0.06
  3. = 0.0576
  4. P(flag) = 0.036 + 0.0576both flag paths
  5. = 0.0936
  6. about 94 in every 1000 bags
Your turn

Question 6 · Excellence

  • Same dog. A bag has been flagged.
  • Find P(it holds food | flagged).
  • Claim: "A flagged bag almost always
  • holds food." Judge the claim.
Question 6 · Excellence
Worked example

Q6: reverse the tree

  • Given flag: flag paths.
  • Path ÷ all flags.
  1. P(F | flag)
  2. = P(F and flag) ÷ P(flag)
  3. = 0.036 ÷ 0.0936food path ÷ all flags
  4. = 0.38464 d.p.
  5. P(flag | F) = 0.9wrong way round
  6. per 1000: 36 food, about 58 not
  7. claim not supported
Your turn

Question 7 · Excellence

  • In a city, 30% of homes have solar
  • panels; 40% of those have a heat pump.
  • Overall, 50% have a heat pump. Claim:
  • "Solar homes are more likely to have one."
Worked example

Q7: the missing branch

  • Solar against no solar.
  • Find the missing branch.
  1. P(S and H) = P(S) × P(H | S)
  2. = 0.3 × 0.4 = 0.12
  3. P(not S and H) = 0.5 − 0.12the rest of P(H)
  4. = 0.38
  5. P(not S) = 1 − 0.3 = 0.7
  6. P(H | not S) = 0.38 ÷ 0.7
  7. = 0.54294 d.p.
  8. 0.4 < 0.5429solar: less likely
  9. claim not supported
Summary

The method that solved all seven

  • Given picks the group, and the group picks the denominator.
StepHow
P(A | B)A given B: B is the group
Tabledivide by that row or column
Venndivide by the given region
Treemultiply along each path
Reversepath ÷ all matching paths
Answernotation, 4 d.p., context

More in AS 91585

Tables, trees and Venn diagrams from written information6 minConditional probability from tables, trees and Venn diagrams6 minIndependent or mutually exclusive? The two tests6 minRelative risk: times as likely, and when you need a ratio6 minTrue probability, model estimates and experimental estimates7 minOrganising information: 7 exam questions solved step by step11 minConditional probability: 7 exam questions solved step by step10 minIndependent or exclusive: 7 exam questions solved step by step10 minRelative risk: 7 exam questions solved step by step10 minProbability estimates: 7 exam questions solved step by step9 min

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