Conditional Probability: The Probability of A Given B

“Given that” shrinks the sample space

A student is chosen at random from $200$. What is the probability they are in the band?

$$P(\text{band}) = \frac{50}{200} = 0.25$$

Now you are told the student plays a sport. What is the probability they are in the band?

	Band	No band	Total
Sport	$18$	$42$	$60$
No sport	$32$	$108$	$140$
Total	$50$	$150$	$200$

Knowing the student plays a sport rules out the $140$ who do not. The sample space is now the $60$ sport players, and $18$ of them are in the band:

$$P(\text{band} \mid \text{sport}) = \frac{18}{60} = 0.30$$

The vertical bar is read “given”. The new information changed the answer from $0.25$ to $0.30$, because it changed which students were in the running.

The formula

Divide the overlap by the size of the condition:

$$P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$$

With the table’s probabilities:

$$P(\text{band} \mid \text{sport}) = \frac{18/200}{60/200} = \frac{18}{60} = 0.30$$

The $200$s cancel, which is why working in counts gives the same answer with less writing: the count in both, over the count in the condition.

Why P(A | B) and P(B | A) are different

Swap the two events:

$$P(\text{sport} \mid \text{band}) = \frac{18}{50} = 0.36$$

The overlap is the same $18$ students. The denominators are not. $P(A \mid B)$ divides by $B$; $P(B \mid A)$ divides by $A$ — different groups, different questions:

Question	Restrict to	Answer
Of sport players, what fraction are in band?	the $60$ sport players	$0.30$
Of band members, what fraction play sport?	the $50$ band members	$0.36$

This distinction matters far beyond tables. A medical test might detect a disease in $99\%$ of people who have it — that is $P(\text{positive} \mid \text{disease})$. A patient who tests positive wants $P(\text{disease} \mid \text{positive})$, and for a rare disease that can be far lower, because most positive results come from the much larger healthy group.

The multiplication rule

Multiply both sides of the formula by $P(B)$:

$$P(A \text{ and } B) = P(B) \times P(A \mid B)$$

It finds the chance of both happening, one step at a time: the chance of the first, times the chance of the second once the first is known.

A bag holds $4$ red and $6$ blue marbles. Two are drawn without replacement. Find $P(\text{both red})$.

$$P(\text{red, then red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$

The $\tfrac{3}{9}$ is the conditional probability: given the first marble was red, $3$ reds remain among $9$ marbles. That is the dependent-events calculation from grade 7, now with its name.

Conditional probability in everyday language

Most everyday statistics are conditional probabilities in disguise. The skill is spotting which event is the condition.

Statement	Written as
”$30\%$ of sport players are in the band”	$P(\text{band} \mid \text{sport}) = 0.30$
“Among smokers, the risk of lung cancer is higher”	$P(\text{cancer} \mid \text{smoker})$
“If it rains, the game is canceled $80\%$ of the time”	$P(\text{canceled} \mid \text{rain}) = 0.80$

The condition is the group being talked about — the word after “among”, “of”, “if” or “given”.

Worked examples

Common mistakes

Practice problems