Practice Maths

L32 — Conditional Probability and Independence

Key Terms

Conditional probability
P(A | B) = P(A ∩ B) ÷ P(B) — the probability of A given that B has already occurred.
Independence
A and B are independent if P(A | B) = P(A), or equivalently P(A ∩ B) = P(A) × P(B).
Multiplication rule (general)
P(A ∩ B) = P(A) × P(B | A) — works for any events, dependent or independent.
With replacement
Each draw is independent — the probability remains unchanged on every draw.
Without replacement
Each draw changes the remaining pool — subsequent draws are dependent on earlier ones.
Bayes' theorem
P(A | B) = P(A) × P(B | A) ÷ P(B) — reverses the conditioning direction.

Conditional probability

The conditional probability of A given B has occurred:

P(A | B) = P(A ∩ B) ÷ P(B), provided P(B) > 0.

Independence

Events A and B are independent if knowing B occurred does not change the probability of A:

P(A | B) = P(A), equivalently P(A ∩ B) = P(A) × P(B).

Multiplication rule (general)

P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)

ScenarioFormula
Conditional probabilityP(A|B) = P(A ∩ B) / P(B)
Independent eventsP(A ∩ B) = P(A) × P(B)
Dependent eventsP(A ∩ B) = P(A) × P(B|A)
Test for independenceP(A|B) = P(A), or P(A ∩ B) = P(A)×P(B)

Sampling with and without replacement

TypeEffect on probability
With replacementDraws are independent — probabilities unchanged
Without replacementDraws are dependent — denominator decreases
P(A) P(A′) P(B|A) A∩B P(B′|A) A∩B′ P(B|A′) A′∩B P(B′|A′) A′∩B′ Multiply along branches to get joint probabilities
Tree diagram: P(A ∩ B) = P(A) × P(B | A)
Hot Tip: Independence means P(A | B) = P(A) — knowing B happened tells you nothing new about A. Check algebraically: if P(A ∩ B) = P(A) × P(B), they are independent. If not, they are dependent.

Worked Example 1 — Conditional probability from a table

Of 50 students: 20 study French (F), 15 study German (G), 8 study both. Find P(G | F).

P(F ∩ G) = 8/50. P(F) = 20/50.

P(G | F) = P(F ∩ G) / P(F) = (8/50) / (20/50) = 8/20 = 2/5.

Interpretation: Given the student studies French, there is a 40% chance they also study German.

Worked Example 2 — Testing independence

P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Are A and B independent?

Test: P(A) × P(B) = 0.4 × 0.5 = 0.2 = P(A ∩ B). ✓

Yes, A and B are independent.

Worked Example 3 — Without replacement (dependent)

A bag has 4 red and 6 blue balls. Two drawn without replacement. Find P(both red).

P(1st red) = 4/10. P(2nd red | 1st red) = 3/9.

P(both red) = (4/10) × (3/9) = 12/90 = 2/15.

Worked Example 4 — Conditional from a Venn diagram

In a class: P(Sport) = 0.6, P(Music) = 0.4, P(both) = 0.25. Find P(Music | Sport).

P(Music | Sport) = P(Music ∩ Sport) / P(Sport) = 0.25 / 0.6 = 5/12 ≈ 0.417.

Worked Example 5 — Medical test (false positives)

A disease affects 1% of the population. A test is 95% accurate for those with the disease and 90% accurate for those without it. Find P(disease | positive test).

P(D) = 0.01, P(D′) = 0.99.

P(+ | D) = 0.95, P(+ | D′) = 0.10 (false positive rate).

P(+) = 0.01×0.95 + 0.99×0.10 = 0.0095 + 0.099 = 0.1085.

P(D | +) = (0.01×0.95) / 0.1085 = 0.0095/0.1085 ≈ 0.0875 ≈ 8.75%.

Even with a positive test, only ~9% chance of actually having the disease — illustrating why rare-disease testing requires careful interpretation.

See Answers ➔

  1. Conditional probability basics. Fluency

    • (a) P(A ∩ B) = 0.12, P(B) = 0.4. Find P(A | B).
    • (b) P(A) = 0.5, P(B) = 0.6, P(A | B) = 0.3. Find P(A ∩ B).
    • (c) A die is rolled. Find P(even | greater than 2).
    • (d) A card is drawn. Find P(King | face card).

  2. Testing independence. Fluency

    • (a) P(A) = 0.3, P(B) = 0.5, P(A ∩ B) = 0.15. Are A and B independent?
    • (b) P(A) = 0.4, P(B) = 0.6, P(A ∩ B) = 0.28. Are they independent?
    • (c) Two coins are tossed. Are the outcomes of the two coins independent?
    • (d) Drawing two cards without replacement. Are the draws independent?

  3. With and without replacement. Fluency

    • (a) Bag: 3 red, 7 blue. Two draws with replacement. Find P(both red).
    • (b) Same bag, two draws without replacement. Find P(both red).
    • (c) Why is the answer to (b) less than (a)?
    • (d) Bag: 5 red, 5 blue. Two draws without replacement. Find P(different colours).

  4. Two-way table and conditional probability. Fluency

    120 people: exercise (Yes/No) × healthy diet (Yes/No):

    Diet YesDiet NoTotal
    Exercise Yes481260
    Exercise No243660
    Total7248120
    • (a) Find P(Diet Yes | Exercise Yes).
    • (b) Find P(Exercise Yes | Diet Yes).
    • (c) Are exercise and diet independent? Test using the formula.
    • (d) Find P(Diet No | Exercise No).

  5. Tree diagram and conditional probability. Understanding

    A box contains 4 red and 6 white balls. Two balls are drawn without replacement. The diagram shows the first and second draws.

    R 4/10 W 6/10 R 3/9 P = 12/90 W 6/9 P = 24/90 R 4/9 P = 24/90 W 5/9 P = 30/90
    • (a) Verify the four branch probabilities sum to 1.
    • (b) Find P(second ball is red).
    • (c) Find P(first ball is red | second ball is red). [Use P(R then R) and P(R then R) + P(W then R).]
    • (d) Are the first and second draws independent? Justify using probabilities.

  6. Probability of dependent events. Understanding

    • (a) In a class of 30, 12 have brown eyes and 8 are left-handed; 4 have both. A student is selected randomly. Find P(brown eyes | left-handed).
    • (b) A box has 5 defective and 15 good items. Two items selected without replacement. Find P(second defective | first defective).
    • (c) P(A) = 0.6, P(B | A) = 0.4, P(B | A′) = 0.2. Find P(A ∩ B) and P(A′ ∩ B).
    • (d) Using (c), find P(B) using the law of total probability.

  7. Independence in context. Understanding

    • (a) A coin is flipped and a die is rolled. Are these events independent? Explain.
    • (b) P(rain tomorrow) = 0.3. P(traffic jam tomorrow) = 0.5. P(both) = 0.2. Are rain and traffic jams independent?
    • (c) In a bag of 10 balls (4 red, 6 blue), one ball is drawn, noted, and replaced. A second is drawn. Find P(red then blue).
    • (d) Same bag but without replacement. Find P(red then blue).

  8. Conditional probability from a Venn diagram. Understanding

    300 survey respondents: 180 own a car (C), 120 use public transport (T), 60 use both.

    • (a) Find P(C | T).
    • (b) Find P(T | C).
    • (c) Are C and T independent?
    • (d) Find P(neither C nor T).

  9. Medical testing. Problem Solving

    A disease affects 2% of the population. A screening test gives a positive result for 95% of those who have the disease, and a positive result for 8% of those who do not have the disease.

    • (a) Define the events D (has disease) and + (positive test). State P(D), P(+|D), P(+|D′).
    • (b) Find P(+) using the law of total probability.
    • (c) Find P(D | +) using Bayes’ theorem: P(D|+) = P(D)×P(+|D) / P(+).
    • (d) Interpret your answer in plain language. Why might this surprise patients?

  10. Quality control. Problem Solving

    A factory has two machines. Machine A produces 60% of items; 3% of Machine A’s items are defective. Machine B produces 40%; 5% are defective.

    • (a) Find P(defective and from Machine A).
    • (b) Find P(defective and from Machine B).
    • (c) Find P(defective).
    • (d) Given an item is defective, find P(it came from Machine A).