Conditional Probability
Learn how to calculate the probability of an event given that another event has already occurred, using tree diagrams and the P(A|B) formula.
The Conditional Probability Formula
Conditional probability answers the question: "What is the probability of event A occurring, given that event B has already occurred?" This is written as P(A|B) and read as "the probability of A given B."
Conditional Probability Formula
P(A|B) = P(A ∩ B)/P(B)
where P(B) > 0
This formula restricts the sample space to only the outcomes in B, then finds the proportion of those that also lie in A.
Rearranging gives the multiplication rule for dependent events: P(A ∩ B) = P(B) × P(A|B). This is essential for tree diagram calculations.
Tree Diagrams
A tree diagram is a visual tool for organising conditional probability problems. Each branch represents an event, and the probability is written along the branch. To find the probability of a path, multiply along the branches.
Example: Drawing Without Replacement
A bag has 3 red and 2 blue marbles. Two are drawn without replacement.
Red (3/5)
R (2/4)
3/5 x 2/4 = 6/20
B (2/4)
3/5 x 2/4 = 6/20
Blue (2/5)
R (3/4)
2/5 x 3/4 = 6/20
B (1/4)
2/5 x 1/4 = 2/20
Independent vs Dependent Events
Two events A and B are independent if knowing that B has occurred does not change the probability of A. Mathematically: P(A|B) = P(A).
Independent
P(A|B) = P(A)
Example: Rolling a die, then flipping a coin.
Dependent
P(A|B) ≠ P(A)
Example: Drawing cards without replacement.
Key Vocabulary
Conditional Probability
The probability of event A occurring given that B has occurred. Written P(A|B).
Tree Diagram
A branching diagram showing all possible outcomes and their probabilities for sequential events.
Dependent Events
Events where the occurrence of one changes the probability of the other.
Without Replacement
Items are not returned after selection, making the sample space smaller with each draw.
Worked Examples
P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find P(A|B).
Step 1: Apply the formula: P(A|B) = P(A ∩ B) / P(B)
Step 2: = 0.3 / 0.5 = 0.6
Answer: P(A|B) = 0.6. Notice P(A|B) = P(A), so A and B are independent.
A box has 4 green and 6 yellow balls. Two are drawn without replacement. Find the probability both are green.
Step 1: P(1st green) = 4/10
Step 2: Given 1st is green, P(2nd green | 1st green) = 3/9
Step 3: P(both green) = 4/10 × 3/9 = 12/90 = 2/15
Answer: P(both green) = 2/15 ≈ 0.133
In a class, 60% study maths, 40% study science, and 25% study both. If a student studies science, what is the probability they also study maths?
Step 1: P(M) = 0.6, P(S) = 0.4, P(M ∩ S) = 0.25
Step 2: P(M|S) = P(M ∩ S) / P(S) = 0.25 / 0.4
Answer: P(M|S) = 0.625
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
P(A ∩ B) = 0.12 and P(B) = 0.4. What is P(A|B)?
Question 2
A bag has 5 red and 3 blue balls. Two are drawn without replacement. What is P(2nd red | 1st red)?
Question 3
If P(A|B) = P(A), what can you conclude about events A and B?
Question 4
P(A) = 0.5, P(B|A) = 0.8. Find P(A ∩ B).
Question 5
A deck of 52 cards. One card is drawn. Given it is a face card (J, Q, K), what is the probability it is a King?
Key Concepts Summary
- ●Conditional probability: P(A|B) = P(A ∩ B) / P(B).
- ●Multiplication rule: P(A ∩ B) = P(B) × P(A|B).
- ●Events are independent if and only if P(A|B) = P(A).
- ●Tree diagrams organise multi-step problems: multiply along branches for joint probabilities.
- ●"Without replacement" creates dependent events because the sample space changes.