The condition is already true.
You are not asking whether a girl was picked. You are told one was — so the boys leave the room.
The denominator is no longer 40. Every conditional probability question is really about which number goes underneath.
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Conditional Probability
“Given that” is an instruction: throw away everyone the condition excludes.
A class of 40 students
24 are girls · 18 take Chemistry · 6 of the Chemistry students are girls
One student is picked at random. Given that the student is a girl, what is the probability she takes Chemistry?
The word "given" shrinks the world to the 24 girls, and 6 of them take Chemistry: 624 = 14. D divides by the whole class of 40; A conditions on Chemistry instead of on girl, which is the reverse question.
Put the four numbers in a table.
Two facts are given, the other two come from subtraction.
| Chemistry | No Chemistry | Total | |
|---|---|---|---|
| Girls | 6 | 18 | 24 |
| Boys | 12 | 4 | 16 |
| Total | 18 | 22 | 40 |
24 − 6 = 18
girls without Chemistry
18 − 6 = 12
boys with Chemistry
16 − 12 = 4
the last corner
Every row and every column has to reach its total. Fill the corner you can, then the rest fall out one subtraction at a time — and the table itself is the check.
Read it off the row.
“Given a girl” means work inside the Girls row and nowhere else.
girls who take Chemistry = 6
the numerator
girls altogether = 24
the new denominator
P(Chem | Girl) = 624 = 14
divide both by 6
The formula says the same thing.
P(A | B) = P(A and B)P(B)
P(Chem and Girl) = 640
out of everybody
P(Girl) = 2440
the condition
6402440 = 624 = 14
the 40s cancel
Which is why the row method works
Both the top and the bottom were counted out of 40, so the 40 disappears. The formula is for when you are handed probabilities instead of head counts — rearranged, it also gives P(A and B) = P(A | B) · P(B).
Two more ways to get the top. If P(A or B) is given instead, the or-rule supplies it: P(A) = 0.3, P(B) = 0.6 and P(A or B) = 0.7 give P(A and B) = 0.3 + 0.6 − 0.7 = 0.2, so P(A | B) = 0.2 ÷ 0.6 = 13. And to turn a conditional around, rebuild the top first: P(B | A) = P(A | B) · P(B) ÷ P(A). With P(A | B) = 12, P(B) = 25 and P(A) = 14, the top is 15 and P(B | A) = (15) ÷ (14) = 45.
Turning it around gives a different answer.
P(Chem | Girl)
6 ÷ 24 = 14
out of the girls
P(Girl | Chem)
6 ÷ 18 = 13
out of the Chemistry students
Same 6 on top. The condition decides the denominator, and the two conditions are not the same group of people.
Read which one comes after the bar
The condition is the second thing named — “given that she is a girl” puts girls underneath. Swapping them is the mistake this topic is built to catch, and both answers will be printed in the choices.
Did the condition actually change anything?
Compare the conditional probability with the plain one.
P(Chem)
18 ÷ 40 = 0.45
any student
P(Chem | Girl)
6 ÷ 24 = 0.25
lower
verdict
0.25 ≠ 0.45
not independent
When P(A | B) equals P(A), knowing B told you nothing and the events are independent — which is exactly when P(A and B) collapses to P(A) · P(B). Here it did tell you something, so that shortcut is not available.
Dice with a condition: shrink the sample space.
sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
the new sample space: 6 outcomes
with a 2: (2, 5) and (5, 2)
2 of them
P = 26 = 13
count inside the condition
Dividing by 36
236 is P(sum 7 and a 2) out of every possible roll. Once the sum is known to be 7, only those six outcomes are left.
Two sources: total probability, then turn it around.
X and late: 0.7 × 0.1 = 0.07
branch X
Y and late: 0.3 × 0.2 = 0.06
branch Y
P(late) = 0.07 + 0.06 = 0.13
add the branches
P(Y | late) = 0.060.13 = 613
Y’s share of the late buses
Think of 100 buses: 70 from X with 7 late, 30 from Y with 6 late. Of the 13 late buses, 6 are from Y.
Answering 0.2 or 0.3
0.2 is P(late | Y) and 0.3 is P(Y) before anything is known. The condition “late” puts 0.13 underneath.
An event that can happen in more than one order.
List the orders that fit, find the probability of each, then add.
3rd test bad, exactly one bad in the first two
the orders are BGB and GBB
BGB: 25 × 34 × 13 = 660
GBB: 35 × 24 × 13 = 660
P = 1260 = 15
On the exam
Expect a two-way table read with a condition, dice with a known sum, P(A | B) from P(A and B), or a two-source story (machines, buses, schools) asking where a defective item came from. Whatever follows “given” goes underneath. Ten-second check: a conditional probability uses a smaller denominator, never the whole sample space.
Check: the table and the rule must agree.
rows
24 + 16 = 40
✓
columns
18 + 22 = 40
✓
multiplication rule
¼ × 2440 = 640
the 6 girls in Chemistry ✓
Now you try
Practice problem
The problem
Of 50 people, 30 own a phone, 20 own a laptop, and 15 own both. Given that a person owns a laptop, find the probability they own a phone.
Given a laptop owner, the world is the 20 laptop owners, and 15 of those also own a phone: 1520 = 34. C divides by all 50. The condition always replaces the denominator — that is the only move in conditional probability.