Probability
Grade 7-9Conditional Probability
Conditional probability asks how likely something is once you already know something else has happened, and that extra piece of information changes the pool you are working from. This lesson covers P(A|B) notation, finding a conditional probability from a Venn diagram or two-way table, using conditional probability to test independence, and combining it with the AND rule.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Understanding P(A|B) notation
P(A|B) means "the probability of A, given that B has already happened". Knowing that B has occurred narrows the group down to only the outcomes where B is true, and the probability of A is then calculated within that smaller group.
P(A|B) = P(A ∩ B) ÷ P(B), since only the outcomes where B happens are being considered, and within those, the proportion where A also happens.
Finding a conditional probability from a Venn diagram
To find P(A|B) from a Venn diagram, ignore everything outside circle B entirely, and find what fraction of circle B is also in circle A.
A Venn diagram shows 10 in Art only, 6 in both Art and Drama, 9 in Drama only. To find P(Art|Drama), only look inside the Drama circle, which has 6 + 9 = 15 students. Of these, 6 are also in Art, giving P(Art|Drama) = 6/15 = 2/5.
Finding a conditional probability from a two-way table
To find a conditional probability from a two-way table, use only the row or column total that matches the condition, rather than the overall total.
A table shows 24 students revised, of whom 21 passed. P(Pass|Revised) = 21/24 = 7/8, using only the total for students who revised.
Testing independence and combining with the AND rule
Two events are independent if knowing one has happened does not change the probability of the other, which means P(A|B) = P(A). If the two values are different, the events are not independent. Rearranging the conditional probability formula also gives P(A and B) = P(A) × P(B|A), useful when a word problem states a conditional probability directly.
If P(A) = 0.4 and P(A|B) = 0.4, the two values are equal, so A and B are independent. If a test has P(pass Test 1) = 0.6 and, given a pass, P(pass Test 2) = 0.75, then P(both) = 0.6 × 0.75 = 0.45.
Worked Examples
Three exam-style questions, fully solved.
P(A ∩ B) = 0.15 and P(B) = 0.5. Find P(A|B).
Easy- 1.Recall the formula: P(A|B) = P(A ∩ B) ÷ P(B)
- 2.Substitute the given values: 0.15 ÷ 0.5
Answer: 0.3
The two-way table shows the exam results of 40 students: of the 24 who revised, 21 passed. Find P(Pass|Revised).
Medium- 1.Identify the total for the given condition: 24 students revised
- 2.Identify how many of those passed: 21
- 3.Write as a fraction: 21/24
Answer: 7/8
The probability that a student passes Test 1 is 0.6. Given that they pass Test 1, the probability they pass Test 2 is 0.75. Find the probability that they pass both tests.
Hard- 1.Recognise this as P(Test 1) × P(Test 2 | Test 1)
- 2.Multiply the two probabilities: 0.6 × 0.75
Answer: 0.45
Avoid These
The most common mistakes students make.
Dividing by the overall total instead of the total for the given condition when finding a conditional probability from a Venn diagram or table.
Confusing P(A|B) with P(B|A), swapping which event is the condition and which is being found.
When testing independence, comparing P(A|B) to the wrong value, or making an arithmetic slip that leads to the wrong conclusion about independence.
Using the conditional probability formula the wrong way round, dividing P(B) by P(A ∩ B) instead of P(A ∩ B) by P(B).
When a word problem gives a conditional probability directly, multiplying the two probabilities in the wrong order or using the wrong pair of values.
FAQ
Questions parents and students ask.
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