Conditional Probability

\( P(A\mid B)=\tfrac{P(A\cap B)}{P(B)}\;(P(B)>0) \)
Probability GCSE

\( P(A)=0.3, P(B)=0.5, P(A∩B)=0.15. Find P(A|B). \)

Hint (H)
\( Use formula P(A|B)=P(A∩B)/P(B). \)

Explanation

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Statement

Conditional probability measures the likelihood of an event \(A\) occurring given that another event \(B\) has already happened. The definition is:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 \]

This means: the probability of \(A\) under the condition that \(B\) is true equals the probability of both \(A\) and \(B\) happening together, divided by the probability of \(B\).

Why it’s true

  • Probabilities are relative frequencies: \(P(A)=\frac{\text{favourable outcomes}}{\text{total outcomes}}\).
  • If we know that \(B\) has occurred, the “new universe” of possible outcomes is reduced to just those in \(B\).
  • The chance that \(A\) also occurs is the fraction of those outcomes in \(B\) that are also in \(A\).
  • This gives the formula \(P(A|B)=P(A\cap B)/P(B)\).

Recipe (how to use it)

  1. Identify the event of interest (\(A\)) and the condition (\(B\)).
  2. Calculate or identify \(P(B)\), the probability of the condition.
  3. Calculate \(P(A \cap B)\), the probability both occur together.
  4. Divide: \(P(A|B)=P(A \cap B)/P(B)\).

Spotting it

Look for wording such as “given that”, “if we already know”, or “conditional on”. Examples: “What is the probability a student passed maths given they passed English?”

Common pairings

  • Two-way tables (joint and conditional probabilities).
  • Tree diagrams (probabilities along branches depend on conditions).
  • Independent events (where \(P(A|B)=P(A)\)).

Mini examples

  1. Card: \(P(\text{red}|\text{heart})=P(\text{red}\cap \text{heart})/P(\text{heart})=1/1=1\).
  2. Die: \(P(\text{even}|\text{>3})=P(\{4,6\})/P(\{4,5,6\})=2/3\).

Pitfalls

  • Forgetting denominator: Always condition by dividing by \(P(B)\).
  • Mixing up \(P(A|B)\) with \(P(B|A)\): They are usually different.
  • Ignoring “given” info: Must restrict the sample space to event \(B\).

Exam strategy

  • Use two-way tables or tree diagrams to organise information.
  • Carefully label events and intersections.
  • Check whether events are independent (then \(P(A|B)=P(A)\)).

Summary

Conditional probability focuses on the chance of one event happening in a restricted situation where another event has already occurred. The formula \(P(A|B)=P(A \cap B)/P(B)\) is essential for probability tables, tree diagrams, and independence tests.

Worked examples

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  1. A card is chosen from a standard deck. Find P(red | hearts).
    1. Event A: red, Event B: hearts.
    2. \( P(A∩B)=P(hearts)=1/4. \)
    3. \( P(B)=1/4. \)
    4. \( P(A|B)=1/4 ÷ 1/4=1. \)
    Answer: 1
  2. Roll a fair die. Find P(even | number >3).
    1. Numbers >3: {4,5,6.
    2. Even in that set: {4,6.
    3. \( P(A|B)=2/3. \)
    Answer: 2/3
  3. \( From a bag with 3 red and 2 blue balls, find P(red | total balls = 5). \)
    1. \( P(red)=3/5. \)
    2. Given info does not change probability.
    3. \( So P(red)=3/5. \)
    Answer: 3/5
  4. A class has 12 boys, 18 girls. 8 girls play football. Find P(football | girl).
    1. Event B: girl (18 students).
    2. \( Event A∩B: football+girl=8. \)
    3. \( P=8/18=4/9. \)
    Answer: 4/9
  5. Two-way table: 20 students, 12 pass maths, 8 fail. 15 pass English, 5 fail. 10 pass both. Find P(pass maths | pass English).
    1. \( Event B: pass English=15. \)
    2. \( Event A∩B: pass both=10. \)
    3. \( P=10/15=2/3. \)
    Answer: 2/3
  6. From a deck, P(king | face card)?
    1. Face cards: 12.
    2. Kings among them: 4.
    3. \( P=4/12=1/3. \)
    Answer: 1/3
  7. \( Roll two dice. Find P(sum=7 | first die=3). \)
    1. \( Condition: first die=3. \)
    2. Possible second outcomes: {1,…,6.
    3. \( Need total 7 → second=4. \)
    4. \( P=1/6. \)
    Answer: 1/6
  8. Bag has 5 red, 3 blue, 2 green. One ball drawn. Find P(red | not green).
    1. \( Total not green=5+3=8. \)
    2. \( Reds among not green=5. \)
    3. \( P=5/8. \)
    Answer: 5/8
  9. \( Medical test: P(positive | disease)=0.95, P(disease)=0.01. If a person tests positive, what is P(disease | positive)? \)
    1. \( Need Bayes: P(D|Pos)=P(Pos|D)P(D)/P(Pos). \)
    2. \( P(Pos)=0.95×0.01+0.05×0.99=0.0095+0.0495=0.059. \)
    3. \( P(D|Pos)=0.0095/0.059≈0.161. \)
    Answer: 0.161
  10. Die rolled. Find P(number>4 | even).
    1. \( Event B: even={2,4,6}. \)
    2. \( Event A∩B: >4 and even={6}. \)
    3. \( P=1/3. \)
    Answer: 1/3