A. Counting and Probability · Probability

Conditional Probability

A1.6About 70 minutes

Learning goals

I can explain what P(B | A), the probability of B given A, means, and find it with a reduced sample space.
I can use the formula P(B | A) = P(A ∩ B) / P(A) and the multiplication rule P(A ∩ B) = P(A) × P(B | A).
I can calculate conditional probabilities from a two-way (contingency) table.
I can use conditional probability to decide whether events are independent, and explain why P(A | B) and P(B | A) are usually different.

Words to know

conditional probability P(B | A)
The probability that B happens, given that A has happened. Read "the probability of B given A".

reduced sample space
The outcomes that remain possible once we know a condition is true; the condition becomes the new sample space.

two-way table (contingency table)
A table that counts data by two categorical variables at once, such as licence (yes/no) and job (yes/no).

multiplication rule
For any events, P(A ∩ B) = P(A) × P(B | A).

When you already know something

Suppose a friend rolls a die, hides it, and tells you, "It's even." What is the probability that it is a 4?

Without the hint, P(4) = 1/6. But the hint rules out 1, 3 and 5. Only {2, 4, 6} remain, still equally likely, so the probability of a 4 is now 1/3. This is a conditional probability:

P(4 | even) = 1/3 (read "the probability of 4 given even").

The condition ("even") becomes a reduced sample space. You count favourable outcomes inside it only.

Knowing the roll is even shrinks the sample space from 6 outcomes to 3. Only one of the three is a 4, so P(4 | even) = 1/3.

The formula

For events A and B with P(A) ≠ 0:

P(B | A) = P(A ∩ B) / P(A)

In words: of all the probability in A, what fraction is also in B? With equally likely outcomes, this is the same as n(A ∩ B) / n(A), the reduced sample space idea.

Rearranging gives the multiplication rule, which you used for draws without replacement in lesson 10:

P(A ∩ B) = P(A) × P(B | A)

Independence, again. If A and B are independent, knowing A does not change B's probability: P(B | A) = P(B). This is another way to test independence, and it says exactly what the word means.

Worked example

Example 1: Two dice

Two dice are rolled. Find (a) P(sum is 8 | first die is 3), (b) P(doubles | sum is 8), (c) P(sum is 8 | doubles).

(a) Given first die 3: reduced space (3, 1), …, (3, 6), 6 outcomes. Sum 8 needs (3, 5): 1 outcome. P = 1/6.
(b) Given sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), 5 outcomes. Doubles: (4, 4). P = 1/5.
(c) Given doubles: 6 outcomes. Sum 8: (4, 4). P = 1/6.
Check (b) with the formula: P(doubles ∩ sum 8) / P(sum 8) = (1/36) / (5/36) = 1/5. ✓

(a) 1/6, (b) 1/5, (c) 1/6. Notice that (b) and (c) are different: P(A | B) is usually not equal to P(B | A).

Worked example

Example 2: Using the formula with given probabilities

At a school, P(a student takes MDM4U) = 0.55 and P(a student takes both MDM4U and MHF4U) = 0.33. Find P(a student takes MHF4U | they take MDM4U).

Let D = takes MDM4U and F = takes MHF4U.
P(F | D) = P(D ∩ F) / P(D) = 0.33 / 0.55 = 0.6.

P(MHF4U | MDM4U) = 0.6: 60% of the MDM4U students also take MHF4U.

Two-way tables

Real data often come in a two-way table (also called a contingency table). The table below shows a survey of 200 Grade 12 students. To find a conditional probability, use only the row or column of the condition as the denominator.

	Job	No job	Total
Licence	60	40	100
No licence	30	70	100
Total	90	110	200

Survey of 200 Grade 12 students (example data): driver's licence and part-time job.

Worked example

Example 3: Reading a two-way table

Use the table. A student is chosen at random. Find (a) P(job), (b) P(job | licence), (c) P(licence | job), (d) Are "has a job" and "has a licence" independent?

(a) P(job) = 90/200 = 0.45.
(b) Condition: licence → use the Licence row (100 students). 60 have a job. P = 60/100 = 0.6.
(c) Condition: job → use the Job column (90 students). 60 have a licence. P = 60/90 = 2/3 ≈ 0.667.
(d) P(job | licence) = 0.6 but P(job) = 0.45. Knowing about the licence changes the probability, so the events are dependent. (Check: P(job ∩ licence) = 60/200 = 0.3, while P(job) × P(licence) = 0.45 × 0.5 = 0.225.)

(a) 0.45, (b) 0.6, (c) 2/3, (d) dependent. Dependence does not prove that a licence causes a job; the unit Two-Variable Data Analysis and Validity looks at cause and effect.

A mosaic of the two-way table: each column is one group of 100 students, split by whether they have a job. The job share is 0.6 in the licence group and 0.3 in the other group, so the events are dependent.

Which number goes on the bottom?

In P(B | A), the condition A is what you already know, so it goes in the denominator. Say it in words first: "Of the students with a licence, what fraction have a job?" The group after "of the" is the denominator.

Worked example

Example 4: The multiplication rule

In a class, 40% of students walk to school. Of those who walk, 25% carry a reusable water bottle. Find P(a student walks and carries a reusable bottle).

Let W = walks, B = carries a bottle. P(W) = 0.4 and P(B | W) = 0.25.
P(W ∩ B) = P(W) × P(B | W) = 0.4 × 0.25 = 0.1.

0.1, or 10% of the class.

Just for fun

Everyday conditional probability: P(you find your keys | you are already late) feels like it is about 0. P(you find your keys | you just paid for a new set) is close to 1.

Hands-on activity

Make your own two-way table

You need: a deck of cards (or ask at least 20 people two yes-or-no questions).

Cards version: shuffle and deal 30 cards. Record each card in a 2 × 2 table: red/black and face card/not face card. Shuffle the whole deck between deals if you want independent draws, or just deal 30 different cards.
From your table, calculate P(face), P(face | red) and P(red | face).
Compare with the full deck values: P(face) = 12/52 ≈ 0.231, P(face | red) = 6/26 ≈ 0.231, P(red | face) = 6/12 = 0.5.
People version: ask people two questions, such as "Do you like winter?" and "Do you skate?" Build the table and decide whether the answers appear independent.

What to look for: in a full deck, colour and "face card" are independent (P(face | red) = P(face)). Your 30-card sample will not match exactly. In the people version, describe the result carefully: a small survey of friends is not a random sample.

Your turn

A die is rolled. Find P(roll is 5 | roll is odd).

A die is rolled. Find P(roll is greater than 2 | roll is even).

One card is drawn. Find P(king | face card) and P(face card | king).

P(A) = 0.5 and P(A ∩ B) = 0.15. Find P(B | A).

P(A) = 0.8 and P(B | A) = 0.35. Find P(A ∩ B).

Two coins are tossed. Find P(both heads | at least one head).

Using the licence and job table, find P(no job | no licence).

More practice: warm-up

How do you read P(B | A)?

In P(B | A), which event goes in the denominator of the formula?

A die shows a number less than 4. P(it is 2)?

If A and B are independent and P(B) = 0.3, what is P(B | A)?

P(A ∩ B) = 0.12, P(A) = 0.4. P(B | A)?

More practice: on your own

Two dice are rolled. Find P(sum ≥ 10 | at least one die shows 6).

A survey of 300 people: 180 drink coffee; 120 drink tea; 60 drink both. Find P(tea | coffee) and P(coffee | tea).

Are drinking coffee and drinking tea independent in the previous question?

Two cards are drawn without replacement. Find P(second is a heart | first is a heart) and P(second is a heart | first is a spade).

Explain why P(A | B) and P(B | A) are usually different, using the example "P(Canadian | lives in Toronto)" and "P(lives in Toronto | Canadian)".

Challenge

Show that if P(B | A) = P(B), then P(A | B) = P(A) (assuming both probabilities are non-zero).

A family has two children. Assume each child is equally likely to be a girl or a boy, independently. Given that at least one is a girl, find P(both are girls). Then find P(both are girls | the older child is a girl).

P(A) = 0.6, P(B) = 0.5, P(A ∪ B) = 0.8. Find P(A | B) and P(B | A′).

Check your understanding

Quiz

A die is rolled. What is P(the roll is 6 | the roll is greater than 3)?

P(A) = 0.6 and P(A ∩ B) = 0.24. What is P(B | A)?

In the licence-and-job table, what is P(licence | no job)?

If A and B are independent with P(A) = 0.3 and P(B) = 0.7, what is P(A | B)?

Two coins are tossed. What is P(two tails | at least one tail)?

Which statement is correct?

For parents and teachers

Big idea (A1.6, conditional probability): P(B | A) is the probability of B once we know A happened. Use A as the new sample space: P(B | A) = P(A ∩ B)/P(A). Events are independent when P(B | A) = P(B).

How to help: Use plain language: "Of the people who have a licence, what fraction have a job?" The phrase after "of the" is the denominator. A two-way table from a quick family survey makes this concrete.

Common difficulties:

Using the wrong denominator (dividing by the total or by the wrong row/column).
Assuming P(A | B) = P(B | A).
Forgetting that in "at least one" conditions, the reduced space has several outcomes (HT, TH and HH).
Concluding that dependence means one thing causes the other.