High School Maths · Year 2 · Unit 13 · Lesson 3 of 5

Conditional Probability: What Changes When You Are Told "Given That"?

About 15 minutes · "given that" and the notation P(A | B) · conditional probability from two-way tables and Venn diagrams · P(A | B) = P(A ∩ B) ÷ P(B) · P(A | B) is not P(B | A) · testing independence with P(A | B) = P(A) · no calculator · exit quiz

Hook

100 people · 40 own a console

ConsoleNo consoleTotal
Adult103040
Child303060
Total4060100

100 people: who owns a games console?

The table shows 100 people. A person who owns a games console is chosen at random. What is the probability that this person is a child?

Only the 40 console owners can be chosen, so divide by 40, not by 100. 30 of them are children, so the probability is 3040 = 34. In Year 1 you chose from one row or column of a two-way table. That is conditional probability: being told something shrinks the list of possible outcomes. This lesson gives it a notation and a formula.

Learn

Master the essentials

P = studies Physics, C = studies Chemistry. 50 students.

Given that

"Given that B has happened" means you only look at the outcomes in B.

Given that a dice shows an even number, only 2, 4 and 6 count.

P(A | B)

Read it as "the probability of A given B". B is the event you are told about.

P(6 | even) = 13

From a two-way table

Divide the cell by the total of the row or column you are given, not by the grand total.

9 of the 15 boys: 915 = 35

From a Venn diagram

Count the overlap, then divide by everyone in the "given" circle.

P(A | B) = n(A ∩ B) ÷ n(B)

The formula

With probabilities instead of counts: P(A | B) = P(A ∩ B) ÷ P(B).

0.12 ÷ 0.4 = 0.3

Order matters

P(A | B) and P(B | A) have different bottoms, so they are usually different.

P(rain | cloudy) is not P(cloudy | rain)

The AND rule again

Rearranging the formula: P(A ∩ B) = P(B) × P(A | B). This works for any events.

0.5 × 0.6 = 0.3

Independence

A and B are independent exactly when P(A | B) = P(A): being told B changes nothing.

P(A) = 0.3 and P(A | B) = 0.3: independent

Worked examples

See it done

1. The table shows the option chosen by each of 110 students. Find (a) P(Music | Year 11), (b) P(Year 11 | Music).

ArtMusicDramaTotal
Year 1018122050
Year 1122241460
Total403634110

The option chosen by each of 110 students.

  1. (a) Given Year 11: only the 60 Year 11 students count. 24 of them chose Music.
  2. P(Music | Year 11) = 2460 = 25.
  3. (b) Given Music: only the 36 Music students count. 24 of them are in Year 11.
  4. P(Year 11 | Music) = 2436 = 23. The two answers are different.

(a) 25 (b) 23

2. The Venn diagram shows how many of 50 students study Physics (P) and Chemistry (C). Find P(P | C) and P(C | P).

  1. Both subjects: 10 students (the overlap).
  2. n(C) = 10 + 6 = 16, so P(P | C) = 1016 = 58.
  3. n(P) = 14 + 10 = 24, so P(C | P) = 1024 = 512.
P = studies Physics, C = studies Chemistry. 50 students.

P(P | C) = 58; P(C | P) = 512

3. P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find P(A | B). Are A and B independent?

  1. Addition rule: P(A ∩ B) = 0.5 + 0.4 − 0.7 = 0.2.
  2. Formula: P(A | B) = 0.2 ÷ 0.4 = 0.5.
  3. P(A | B) = 0.5 = P(A): being told B does not change the probability of A.
  4. So A and B are independent.

P(A | B) = 0.5; yes, they are independent

Common mistake: Dividing by the grand total instead of the total of the "given" group. In P(A | B), B is what you are told, so n(B) or P(B) goes on the bottom. Also, P(A | B) and P(B | A) are usually different: check which event comes after the bar.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Find the group you are told about first: that total goes on the bottom. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.A fair dice is rolled. Given that the score is greater than 2, find the probability that it is even.

Not quite. This divides by all 6 scores. Given "greater than 2", only 3, 4, 5 and 6 count.

Solution

  1. Given greater than 2: the outcomes are 3, 4, 5 and 6.
  2. The even ones are 4 and 6.
  3. P = 24 = 12.

Answer: B. 12.

Correct.

Solution

  1. Given greater than 2: the outcomes are 3, 4, 5 and 6.
  2. The even ones are 4 and 6.
  3. P = 24 = 12.

Not quite. This includes 2. "Greater than 2" leaves it out.

Solution

  1. Given greater than 2: the outcomes are 3, 4, 5 and 6.
  2. The even ones are 4 and 6.
  3. P = 24 = 12.

Answer: B. 12.

Not quite. This is P(greater than 2).

Solution

  1. Given greater than 2: the outcomes are 3, 4, 5 and 6.
  2. The even ones are 4 and 6.
  3. P = 24 = 12.

Answer: B. 12.

2.The Venn diagram shows the number of people in each region. Find P(B | A).

The number of people in each region.

Not quite. This divides the overlap by everyone, 40. Given A, divide by n(A).

Solution

  1. Given A: n(A) = 12 + 8 = 20.
  2. In both A and B: 8.
  3. P(B | A) = 820 = 25.

Answer: B. 25.

Correct.

Solution

  1. Given A: n(A) = 12 + 8 = 20.
  2. In both A and B: 8.
  3. P(B | A) = 820 = 25.

Not quite. This divides by n(B). That is P(A | B).

Solution

  1. Given A: n(A) = 12 + 8 = 20.
  2. In both A and B: 8.
  3. P(B | A) = 820 = 25.

Answer: B. 25.

Not quite. This divides by n(A ∪ B).

Solution

  1. Given A: n(A) = 12 + 8 = 20.
  2. In both A and B: 8.
  3. P(B | A) = 820 = 25.

Answer: B. 25.

3.P(A ∩ B) = 0.18 and P(B) = 0.6. Find P(A | B).

Not quite. This multiplies. P(A | B) = P(A ∩ B) ÷ P(B).

Solution

  1. P(A | B) = P(A ∩ B) ÷ P(B).
  2. = 0.18 ÷ 0.6 = 1.8 ÷ 6 = 0.3.

Answer: D. 0.3.

Not quite. This subtracts. Divide instead.

Solution

  1. P(A | B) = P(A ∩ B) ÷ P(B).
  2. = 0.18 ÷ 0.6 = 1.8 ÷ 6 = 0.3.

Answer: D. 0.3.

Not quite. This divides the wrong way round. A probability cannot be more than 1.

Solution

  1. P(A | B) = P(A ∩ B) ÷ P(B).
  2. = 0.18 ÷ 0.6 = 1.8 ÷ 6 = 0.3.

Answer: D. 0.3.

Correct.

Solution

  1. P(A | B) = P(A ∩ B) ÷ P(B).
  2. = 0.18 ÷ 0.6 = 1.8 ÷ 6 = 0.3.

Exam standard (Several steps, like real exam questions.)

4.The table shows how 90 employees travel to work. An employee who cycles is chosen at random. Find the probability that this employee is under 30.

CarBusCycleTotal
Under 3014101640
30 and over2618650
Total40282290

How 90 employees travel to work.

Correct.

Solution

  1. Given "cycles": the Cycle column has 16 + 6 = 22 employees.
  2. 16 of them are under 30.
  3. P = 1622 = 811.

Not quite. This is P(cycles | under 30): 16 out of the 40 under 30.

Solution

  1. Given "cycles": the Cycle column has 16 + 6 = 22 employees.
  2. 16 of them are under 30.
  3. P = 1622 = 811.

Answer: A. 811.

Not quite. This divides by all 90 employees. Only the cyclists can be chosen.

Solution

  1. Given "cycles": the Cycle column has 16 + 6 = 22 employees.
  2. 16 of them are under 30.
  3. P = 1622 = 811.

Answer: A. 811.

Not quite. This is P(cycles) for an employee chosen from all 90.

Solution

  1. Given "cycles": the Cycle column has 16 + 6 = 22 employees.
  2. 16 of them are under 30.
  3. P = 1622 = 811.

Answer: A. 811.

5.P(A) = 0.45, P(B) = 0.3 and P(A ∩ B) = 0.15. Which working gives P(B | A)?

Correct.

Solution

  1. Given A, so P(A) goes on the bottom.
  2. P(B | A) = P(A ∩ B) ÷ P(A) = 0.15 ÷ 0.45 = 13.

Not quite. This divides by P(B). That gives P(A | B).

Solution

  1. Given A, so P(A) goes on the bottom.
  2. P(B | A) = P(A ∩ B) ÷ P(A) = 0.15 ÷ 0.45 = 13.

Answer: A. 0.15 ÷ 0.45.

Not quite. This multiplies P(A) by P(B). That is only P(A ∩ B) for independent events.

Solution

  1. Given A, so P(A) goes on the bottom.
  2. P(B | A) = P(A ∩ B) ÷ P(A) = 0.15 ÷ 0.45 = 13.

Answer: A. 0.15 ÷ 0.45.

Not quite. This divides P(B) by P(A). Use the overlap, P(A ∩ B), on top.

Solution

  1. Given A, so P(A) goes on the bottom.
  2. P(B | A) = P(A ∩ B) ÷ P(A) = 0.15 ÷ 0.45 = 13.

Answer: A. 0.15 ÷ 0.45.

6.P(A) = 0.5, P(B) = 0.4 and P(A | B) = 0.35. Find P(A ∪ B).

Not quite. This adds P(A) and P(B) and forgets to take away the overlap.

Solution

  1. P(A ∩ B) = P(B) × P(A | B) = 0.4 × 0.35 = 0.14.
  2. Addition rule: P(A ∪ B) = 0.5 + 0.4 − 0.14.
  3. = 0.76.

Answer: C. 0.76.

Not quite. This takes away P(A | B). Take away P(A ∩ B) = 0.4 × 0.35.

Solution

  1. P(A ∩ B) = P(B) × P(A | B) = 0.4 × 0.35 = 0.14.
  2. Addition rule: P(A ∪ B) = 0.5 + 0.4 − 0.14.
  3. = 0.76.

Answer: C. 0.76.

Correct.

Solution

  1. P(A ∩ B) = P(B) × P(A | B) = 0.4 × 0.35 = 0.14.
  2. Addition rule: P(A ∪ B) = 0.5 + 0.4 − 0.14.
  3. = 0.76.

Not quite. This is P(A ∩ B).

Solution

  1. P(A ∩ B) = P(B) × P(A | B) = 0.4 × 0.35 = 0.14.
  2. Addition rule: P(A ∪ B) = 0.5 + 0.4 − 0.14.
  3. = 0.76.

Answer: C. 0.76.

7.In a sports club, the probability that a member swims is 0.55, the probability that a member runs is 0.4, and the probability that a member swims or runs is 0.75. A member who runs is chosen at random. Find the probability that this member swims.

Not quite. This is P(swims and runs). Divide by P(runs).

Solution

  1. P(both) = 0.55 + 0.4 − 0.75 = 0.2.
  2. P(swims | runs) = 0.2 ÷ 0.4.
  3. = 0.5.

Answer: B. 0.5.

Correct.

Solution

  1. P(both) = 0.55 + 0.4 − 0.75 = 0.2.
  2. P(swims | runs) = 0.2 ÷ 0.4.
  3. = 0.5.

Not quite. This divides by P(swims). That is P(runs | swims).

Solution

  1. P(both) = 0.55 + 0.4 − 0.75 = 0.2.
  2. P(swims | runs) = 0.2 ÷ 0.4.
  3. = 0.5.

Answer: B. 0.5.

Not quite. This divides by P(swims or runs). Given "runs", divide by P(runs).

Solution

  1. P(both) = 0.55 + 0.4 − 0.75 = 0.2.
  2. P(swims | runs) = 0.2 ÷ 0.4.
  3. = 0.5.

Answer: B. 0.5.

Challenge (Harder problems that mix ideas or need reasoning.)

8.The Venn diagram shows the number of people in each region. P(B | A) = 37. A person is chosen at random. Find P(A).

The number of people in each region.

Correct.

Solution

  1. n(A) = x + 6, so 6x + 6 = 37.
  2. 3(x + 6) = 42, so x + 6 = 14 and x = 8.
  3. n(A) = 14 and n(U) = 8 + 6 + 9 + 5 = 28.
  4. P(A) = 1428 = 12.

Not quite. This uses A only. The 6 people in the overlap are in A too.

Solution

  1. n(A) = x + 6, so 6x + 6 = 37.
  2. 3(x + 6) = 42, so x + 6 = 14 and x = 8.
  3. n(A) = 14 and n(U) = 8 + 6 + 9 + 5 = 28.
  4. P(A) = 1428 = 12.

Answer: A. 12.

Not quite. This is P(B | A), the value you were given.

Solution

  1. n(A) = x + 6, so 6x + 6 = 37.
  2. 3(x + 6) = 42, so x + 6 = 14 and x = 8.
  3. n(A) = 14 and n(U) = 8 + 6 + 9 + 5 = 28.
  4. P(A) = 1428 = 12.

Answer: A. 12.

Not quite. This solves 6 ÷ x = 37. But n(A) = x + 6.

Solution

  1. n(A) = x + 6, so 6x + 6 = 37.
  2. 3(x + 6) = 42, so x + 6 = 14 and x = 8.
  3. n(A) = 14 and n(U) = 8 + 6 + 9 + 5 = 28.
  4. P(A) = 1428 = 12.

Answer: A. 12.

9.P(A) = 0.7, P(B) = 0.4 and P(A ∩ B) = 0.3. Find P(B | A′).

Not quite. This is P(A′ ∩ B). Divide it by P(A′).

Solution

  1. P(A′ ∩ B) is B but not A: 0.4 − 0.3 = 0.1.
  2. P(A′) = 1 − 0.7 = 0.3.
  3. P(B | A′) = 0.1 ÷ 0.3 = 13.

Answer: D. 13.

Not quite. This divides by P(B). That is P(A′ | B).

Solution

  1. P(A′ ∩ B) is B but not A: 0.4 − 0.3 = 0.1.
  2. P(A′) = 1 − 0.7 = 0.3.
  3. P(B | A′) = 0.1 ÷ 0.3 = 13.

Answer: D. 13.

Not quite. This is P(B | A), not P(B | A′).

Solution

  1. P(A′ ∩ B) is B but not A: 0.4 − 0.3 = 0.1.
  2. P(A′) = 1 − 0.7 = 0.3.
  3. P(B | A′) = 0.1 ÷ 0.3 = 13.

Answer: D. 13.

Correct.

Solution

  1. P(A′ ∩ B) is B but not A: 0.4 − 0.3 = 0.1.
  2. P(A′) = 1 − 0.7 = 0.3.
  3. P(B | A′) = 0.1 ÷ 0.3 = 13.

10.A card is taken at random from cards numbered 1 to 12. A is "an even number" and B is "a multiple of 3". Which statement is true?

Not quite. This divides by all 12 cards. Given B, only 3, 6, 9 and 12 count.

Solution

  1. Given B: the cards are 3, 6, 9 and 12. The even ones are 6 and 12.
  2. P(A | B) = 24 = 12.
  3. P(A) = 612 = 12. They are equal, so A and B are independent.

Answer: C. P(A | B) = 12 = P(A), so A and B are independent.

Not quite. Sharing outcomes means they are not mutually exclusive. Independent events can share outcomes.

Solution

  1. Given B: the cards are 3, 6, 9 and 12. The even ones are 6 and 12.
  2. P(A | B) = 24 = 12.
  3. P(A) = 612 = 12. They are equal, so A and B are independent.

Answer: C. P(A | B) = 12 = P(A), so A and B are independent.

Correct.

Solution

  1. Given B: the cards are 3, 6, 9 and 12. The even ones are 6 and 12.
  2. P(A | B) = 24 = 12.
  3. P(A) = 612 = 12. They are equal, so A and B are independent.

Not quite. That is P(A or B). It is not a test for independence.

Solution

  1. Given B: the cards are 3, 6, 9 and 12. The even ones are 6 and 12.
  2. P(A | B) = 24 = 12.
  3. P(A) = 612 = 12. They are equal, so A and B are independent.

Answer: C. P(A | B) = 12 = P(A), so A and B are independent.

Apply

Café customers

A café records what 150 customers bought. The table shows the results.

Bought foodNo foodTotal
Hot drink483280
Cold drink274370
Total7575150

What 150 café customers bought.

1. A customer who bought a hot drink is chosen at random. Find the probability that this customer bought food.

2. A customer who bought food is chosen at random. Find the probability that this customer bought a hot drink.

3. Are "buys a hot drink" and "buys food" independent? Explain using your answers.

  1. Given hot drink: 80 customers, 48 bought food. P = 4880 = 35.
  2. Given food: 75 customers, 48 bought a hot drink. P = 4875 = 1625.
  3. P(food) = 75150 = 12, but P(food | hot drink) = 35. They are different, so the events are not independent: hot-drink customers are more likely to buy food.

Next: the Lesson 3 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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