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Conditional probability: change the reference group

Final-year concepts · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Fractions: share the same wholeProbability: read a tree
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

P(B given A) measures B within A when P(A)>0, using P(A∩B)/P(A). The denominator is the condition’s probability. Reversing condition and event usually changes the answer. Independence means conditioning on A does not change B’s probability where defined.

p=0.12/0.3
THE PICTURE TO REMEMBERConditional probability: change the reference group
  1. 01Condition
  2. 02Intersection
  3. 03Divide by condition
  4. 04Do not reverse

WORKED EXAMPLE

From principle to reasoning

Among 100 people, 30 are in A and 12 in both. P(B|A)=12/30=0.4. If B contains 20, P(A|B)=12/20=0.6: same overlap, different reference group.

Solve a problem in this subject

  1. Define the objects

    State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.

  2. Choose a transformation

    Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.

  3. Show the reasoning

    Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.

  4. Test and conclude

    Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.

YOUR TURN

01 · Try before checking

P(A)=0.5 and P(A∩B)=0.2. Find P(B|A).

One hint

The condition determines the denominator.

Reveal and understand the solution

0.2/0.5=0.4. Finding P(A|B) additionally needs P(B).

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

P(A) = 0.4 and P(A ∩ B) = 0.1. Find P(B | A).

A hint to get started

Restrict the population to outcomes where A occurs.

Step-by-step solution
  1. P(B
  2. A) = P(A ∩ B)/P(A).
  3. 0.1/0.4 = 0.25.
  4. The denominator is P(A), not P(B).

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

P(A) = 0.3, P(B | A) = 0.8, P(B | not A) = 0.2. Find P(B).

A hint to get started

Add the two disjoint paths to B.

Step-by-step solution
  1. P(A ∩ B) = 0.3 × 0.8 = 0.24.
  2. P(not A ∩ B) = 0.7 × 0.2 = 0.14.
  3. P(B) = 0.38.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

Using the preceding data, find P(A | B). Is it P(B | A)?

A hint to get started

Use intersection 0.24 and total P(B) = 0.38.

Step-by-step solution
  1. P(A
  2. B) = 0.24/0.38 = 12/19 ≈ 0.632.
  3. P(B
  4. A) is 0.8.
  5. Reversing conditioning changes the reference population.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Conditional probability is not an intersection. Do not assume independence.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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