Conditional Probability

SPEC 3.1, 3.2 • PROBABILITY

Understanding probability when we have additional information

What is Conditional Probability?

Conditional probability is the probability of an event occurring, given that another event has already occurred.

Key Concept

When we know something has happened, it changes our sample space. We're no longer considering all possible outcomes—only those consistent with what we know.

Example: What's the probability of rolling a 6 on a die?

Normally: P(6) = 1/6

But if someone tells you "the number is even", now: P(6 | even) = 1/3

The sample space reduced from {1,2,3,4,5,6} to just {2,4,6}

The Idea

Conditional probability answers: "Given what I now know, what's the probability?"

Notation & Formula

Notation

P(A|B) is read as "the probability of A given B"

It means: the probability that A occurs, knowing that B has occurred.

Formula

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

where P(B) ≠ 0

Numerator

P(A ∩ B) = probability that BOTH A and B occur

Denominator

P(B) = probability of the "given" event (our new sample space)

Memory Tip

The event after the "|" goes in the denominator. Think: "given B" means "divide by P(B)"

Worked Example01

Events A and B are such that P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2. Find P(A|B).

1

Write the formula:

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

2

Substitute the values:

P (A ∣ B) = \frac{0.2}{0.5}

3

Calculate:

P (A ∣ B) = 0.4

Answer: P(A|B) = 0.4

Worked Example02

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

1

From conditional probability:

P (A \cap B) = P (A ∣ B) \times P (B) = 0.5 \times P (B)

2

Using addition rule:

P (A \cup B) = P (A) + P (B) - P (A \cap B)

3

Substitute:

0.8 = 0.6 + P (B) - 0.5 \times P (B)

4

Simplify:

0.8 = 0.6 + 0.5 \times P (B)

5

Solve:

0.2 = 0.5 \times P (B)

, so

P (B) = 0.4

Answer: P(B) = 0.4

From Venn Diagrams

When working with Venn diagrams, conditional probability means restricting to one circle and finding what fraction is in the overlap.

P(A|B): Given B occurred

■ Denominator (B) ■ Numerator (A∩B)

P(B|A): Given A occurred

■ Denominator (A) ■ Numerator (A∩B)

Key Insight

The "given" event becomes your whole sample space. Green = denominator (the whole "given" circle). Orange = numerator (the intersection we want).

03Worked Example

In a Venn diagram: P(A only) = 0.3, P(B only) = 0.25, P(A ∩ B) = 0.15, P(neither) = 0.3. Find P(A|B).

1

Find P(B): P(B) = P(B only) + P(A ∩ B) = 0.25 + 0.15 = 0.4

2

We have P(A ∩ B) = 0.15

3

Apply formula:

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.15}{0.4}

4

Calculate:

P (A ∣ B) = 0.375

Answer: P(A|B) = 0.375 or 3/8

Worked Example04

Events A and B satisfy P(A) = 0.5, P(B) = 0.6, and P(A' ∩ B') = 0.2. Find P(B|A) and P(A|B').

1

From P(A' ∩ B') = 0.2: This is "neither", so P(A ∪ B) = 1 − 0.2 = 0.8

2

Using addition rule: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.6 − 0.8 = 0.3

3

For P(B|A):

P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{0.3}{0.5} = 0.6

4

For P(A|B'): First find P(A ∩ B') = P(A) − P(A ∩ B) = 0.5 − 0.3 = 0.2

5

And P(B') = 1 − P(B) = 1 − 0.6 = 0.4

6

So:

P (A ∣ B^{'}) = \frac{P ( A \cap B ^{'} )}{P ( B ^{'} )} = \frac{0.2}{0.4} = 0.5

Answer: P(B|A) = 0.6, P(A|B') = 0.5

From Two-Way Tables

Two-way tables are excellent for finding conditional probabilities. The "given" condition tells you which row or column total to use as your denominator.

PassFailTotal

Male451560

Female35540

Total8020100

P(Pass | Male)

Restrict to Male row: 45/60 = 0.75

Row total is denominator

P(Male | Pass)

Restrict to Pass column: 45/80 = 0.5625

Column total is denominator

Method

Find the row/column for the "given" event → its total is your denominator.

05Worked Example

A survey of 200 students: 80 study Maths, 60 study Physics, 30 study both. Find P(Physics | Maths) and P(Maths | Physics).

PhysicsNot PhysicsTotal

Maths305080

Not Maths3090120

Total60140200

1

P(Physics | Maths): Restrict to Maths row → 30/80 = 0.375

2

P(Maths | Physics): Restrict to Physics column → 30/60 = 0.5

Answer: P(Physics | Maths) = 0.375, P(Maths | Physics) = 0.5

Note: These are different! P(A|B) ≠ P(B|A)

Rearranging the Formula

The conditional probability formula can be rearranged to find the intersection probability:

Multiplication Rule

P (A \cap B) = P (A ∣ B) \times P (B)

Alternative Form

We can also write:

P (A \cap B) = P (B ∣ A) \times P (A)

Both give the same result since intersection is commutative.

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

P (A \cap B) = P (A ∣ B) \times P (B)

When to Use

Use the multiplication rule when you know conditional probability and need to find the intersection.

Worked Example06

Given that P(A) = 0.6, P(B) = 0.5, and P(A|B) = 0.4, find P(A ∪ B).

1

First find P(A ∩ B) using multiplication rule:

•

P (A \cap B) = P (A ∣ B) \times P (B) = 0.4 \times 0.5 = 0.2

2

Now use addition rule:

•

P (A \cup B) = P (A) + P (B) - P (A \cap B)

3

Substitute:

P (A \cup B) = 0.6 + 0.5 - 0.2 = 0.9

Answer: P(A ∪ B) = 0.9

Worked Example07

Events A and B satisfy P(B|A) = 0.4, P(B|A') = 0.7, and P(A) = 0.6. Find P(B) and P(A|B).

1

Use total probability: P(B) = P(B|A)P(A) + P(B|A')P(A')

2

P(A') = 1 − 0.6 = 0.4

3

P(B) = 0.4 × 0.6 + 0.7 × 0.4 = 0.24 + 0.28 = 0.52

4

For P(A|B): First find P(A ∩ B) = P(B|A) × P(A) = 0.4 × 0.6 = 0.24

5

Then:

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.24}{0.52} = \frac{6}{13} \approx 0.462

Answer: P(B) = 0.52, P(A|B) = 6/13 ≈ 0.462

Independence Revisited

Conditional probability gives us a new way to understand and test for independence.

Independence via Conditional Probability

Events A and B are independent if and only if:

P (A ∣ B) = P (A)

Knowing B occurred doesn't change the probability of A

If Independent

P(A|B) = P(A)

P(B|A) = P(B)

P(A ∩ B) = P(A) × P(B)

If NOT Independent

P(A|B) ≠ P(A)

Knowing B changes probability of A

Events are "dependent"

Intuition

Independence means "B happening tells us nothing new about A"

Worked Example08

Events A and B satisfy P(A) = 0.3, P(B) = 0.5, and P(A ∩ B) = 0.15. Determine whether A and B are independent.

1

Calculate P(A|B):

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.15}{0.5} = 0.3

2

Compare with P(A): P(A) = 0.3

3

Since P(A|B) = P(A) = 0.3, the events ARE independent.

Answer: A and B are independent

Verification: P(A) × P(B) = 0.3 × 0.5 = 0.15 = P(A ∩ B) ✓

Common Mistakes

Critical Warning

P (A ∣ B) \neq = P (B ∣ A)

in general!

The order matters. These are different probabilities with different denominators.

P(A|B) = (A∩B) / B

P(B|A) = (A∩B) / A

Example: In a class, 40% are girls. Of the girls, 25% wear glasses. Overall 20% wear glasses.

P(Glasses | Girl) = 0.25 but P(Girl | Glasses) = (0.4 × 0.25)/0.2 = 0.5

Worked Example09

For events A and B: P(A) = 0.4, P(B) = 0.6, P(A ∩ B) = 0.24. Calculate P(A|B) and P(B|A). Are A and B independent?

1

Calculate P(A|B):

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.24}{0.6} = 0.4

2

Calculate P(B|A):

P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{0.24}{0.4} = 0.6

3

Compare: P(A|B) = 0.4 = P(A), so events ARE independent

4

Verify: P(B|A) = 0.6 = P(B) ✓

Answer: P(A|B) = 0.4, P(B|A) = 0.6, and A and B are independent

10

Worked Example

2% of a population has a disease. A test correctly identifies 95% of those with the disease and 90% of those without. If a person tests positive, what is the probability they have the disease?

1

Let D = disease, + = positive. Given: P(D) = 0.02, P(+|D) = 0.95, P(−|D') = 0.90

2

So P(+|D') = 0.10 (false positive rate)

3

P(+) = P(+|D)P(D) + P(+|D')P(D') = 0.95×0.02 + 0.10×0.98 = 0.117

4

P (D ∣ +) = \frac{P ( + ∣ D ) \times P ( D )}{P ( + )} = \frac{0.019}{0.117} \approx 0.162

Answer: P(Disease | Positive) ≈ 0.162 or about 16%

Surprising! Despite a "95% accurate" test, a positive result only means 16% chance of disease.

Practice

Question 1: Events A and B satisfy P(A) = 0.5, P(B) = 0.4, and P(A ∩ B) = 0.1. Find:

(a) P(A|B) (b) P(B|A) (c) Are A and B independent?

(a)

P (A ∣ B) = \frac{0.1}{0.4} = 0.25

(b)

P (B ∣ A) = \frac{0.1}{0.5} = 0.2

(c) P(A|B) = 0.25 ≠ P(A) = 0.5, so A and B are NOT independent

Practice

Question 2: In a Venn diagram for events A and B:

P(A only) = 0.2, P(A ∩ B) = 0.3, P(B only) = 0.1, P(neither) = 0.4

Find P(B|A) and P(A|B').

P(B|A): P(A) = 0.2 + 0.3 = 0.5, so P(B|A) = 0.3/0.5 = 0.6

P(A|B'): P(A ∩ B') = P(A only) = 0.2, P(B') = 0.2 + 0.4 = 0.6

So P(A|B') = 0.2/0.6 = 1/3 ≈ 0.333

Practice

Question 3: 60% of balls in a bag are large, 40% are small. Of large balls, 30% are red. Of small balls, 50% are red. A ball drawn at random is red. Find P(Large | Red).

Step 1: P(Red) = P(Red|L)P(L) + P(Red|S)P(S) = 0.3×0.6 + 0.5×0.4 = 0.38

Step 2: P(Red ∩ Large) = P(Red|L) × P(L) = 0.3 × 0.6 = 0.18

Step 3: P(Large|Red) = 0.18/0.38 = 9/19 ≈ 0.474

Summary

Key Formulas

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

P (A \cap B) = P (A ∣ B) \times P (B)

Independence Test

A and B independent ⟺ P(A|B) = P(A)

Key Warning

P(A|B) ≠ P(B|A) in general!

Exam Tips

Event after "|" goes in the denominator

For Venn diagrams: the "given" circle is your 100%

For tables: row/column total of "given" is your denominator

Always check: have you answered P(A|B) or P(B|A)?