Conditional Probability

What 'given that' does to the question

A conditional probability is the probability of one event when you already know that another has happened. The phrases that signal it are given that, if, and of the people who: 'the probability that the second counter is red given that the first was blue'.

Nothing about the world changes. What changes is the group you are choosing from. Knowing that the first counter was blue rules out some of the possibilities, so the sample space shrinks to the outcomes that match what you were told. The denominator of the fraction is the size of that smaller group, and the numerator counts the outcomes inside it that also satisfy the event you want.

Two fair dice show the shrinking clearly. The sample space has 36 outcomes and $P(\text{total is }8)=\frac{5}{36}$. Now suppose you are told the first dice shows a 4. Only the six outcomes (4,1) to (4,6) are still possible, and just one of them, (4,4), gives a total of 8. So $P(\text{total is }8\text{ given the first dice is }4)=\frac{1}{6}$, not $\frac{5}{36}$.

The order matters. $P(A\text{ given }B)$ and $P(B\text{ given }A)$ are different questions with different denominators: whatever is named after 'given that' is the group you divide by.

• Told that the first dice is a 4, the 36 outcomes fall to 6
• $P(\text{total }8)=\frac{5}{36}$, but $P(\text{total }8\text{ given first is }4)=\frac{1}{6}$
• The event named after 'given that' supplies the denominator
• The numerator is the count of the outcomes in that smaller group that satisfy the event you want

Reading it off a two-way table

A two-way table is the friendliest place to find a conditional probability, because the group you were given is already a row total or a column total. In this year group of 60 students, 30 are boys and 30 are girls, 18 of the boys and 6 of the girls walk to school, and the rest take the bus.

'Of the students who walk to school, one is chosen at random. Find the probability that the student is a boy.' The given group is the 24 who walk, so the denominator is 24. Of those 24, 18 are boys, so $P(\text{boy given walks})=\frac{18}{24}=\frac{3}{4}$.

Swap the given group and the answer changes. $P(\text{walks given boy})=\frac{18}{30}=\frac{3}{5}$, because the denominator is now the 30 boys. The numerator is the same 18 either way; only the group named after 'given that' moves.

The two conditional probabilities for one given group always add to 1. Every student who walks is either a boy or a girl, so $\frac{18}{24}+\frac{6}{24}=1$. That is a free check on your reading of the table.

• $P(\text{boy given walks})=\frac{18}{24}=\frac{3}{4}$
• $P(\text{girl given walks})=\frac{6}{24}=\frac{1}{4}$, and the two add to 1
• $P(\text{walks given boy})=\frac{18}{30}=\frac{3}{5}$
• $P(\text{bus given girl})=\frac{24}{30}=\frac{4}{5}$

Reading it off a Venn diagram

On a Venn diagram the given event is one whole circle, overlap included. The denominator is everything inside that circle; the numerator is the part of the overlap that also satisfies the event you want.

In a survey of 50 people, 30 own a cat, 20 own a dog and 12 own both. The four regions are 18 cat only, 12 both, 8 dog only, and 12 who own neither.

Given that a person owns a dog, the group is the 20 inside the dog circle - the 8 dog only plus the 12 who own both. So $P(\text{cat given dog})=\frac{12}{20}=\frac{3}{5}$.

Reverse it and the circle changes: $P(\text{dog given cat})=\frac{12}{30}=\frac{2}{5}$. The same 12 sits on top both times, because both questions are about the overlap; only the circle underneath moves.

Tree diagrams: the branches already are conditional

On a tree diagram the second level of branches is a set of conditional probabilities. Each one answers 'given that this first outcome happened, what now?'. That is exactly why the numbers on the second level are unequal when the first counter is not replaced.

A bag holds 3 red and 2 blue counters, and two are taken without replacement. The first pair of branches is $\frac{3}{5}$ and $\frac{2}{5}$. After a red, four counters remain with 2 red and 2 blue, so the second pair is $\frac{2}{4}$ and $\frac{2}{4}$. After a blue, four counters remain with 3 red and 1 blue, so that pair is $\frac{3}{4}$ and $\frac{1}{4}$.

Reading a conditional probability off the tree is a one-step job. $P(\text{the second counter is red given the first was blue})=\frac{3}{4}$, straight off the branch that follows blue. To find the probability of both events, multiply along the path: $P(\text{the first is blue and the second is red})=\frac{2}{5}\times \frac{3}{4}=\frac{6}{20}=\frac{3}{10}$.

Notice how the two answers differ. $\frac{3}{4}$ is a single branch and is conditional; $\frac{3}{10}$ is the product of two branches and is not. A question that says 'given that' is asking for the branch, and multiplying there loses the mark.

The multiplication rule, rearranged

Multiplying along a tree path gives $P(A\text{ and }B)=P(A)\times P(B\text{ given }A)$. The second factor is conditional because it is the probability of B in the world where A has already happened.

Rearranged, the same rule turns a joint probability into a conditional one: $P(B\text{ given }A)=\frac{P(A\text{ and }B)}{P(A)}$. Divide the probability of both events by the probability of the one you were given.

The cat and dog survey shows it working. $P(\text{cat and dog})=\frac{12}{50}$ and $P(\text{dog})=\frac{20}{50}$, so $P(\text{cat given dog})=\frac{12}{50}÷\frac{20}{50}=\frac{12}{20}=\frac{3}{5}$, which is the answer the Venn diagram gave by counting.

The rule also explains why the denominator is the given event. $P(A)$ is the fraction of the world in which A happened, and dividing by it rescales the world so that A is certain. Everything you can find by counting regions follows from that one line.

• $P(A\text{ and }B)=P(A)\times P(B\text{ given }A)$
• $P(B\text{ given }A)=\frac{P(A\text{ and }B)}{P(A)}$
• Two dice: $P(\text{total }8\text{ and first is }4)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$, and $\frac{1}{36}÷\frac{1}{6}=\frac{1}{6}$ gives back the conditional probability found by counting

Independence: when knowing one event changes nothing

Two events are independent when being told that one has happened leaves the probability of the other exactly as it was: $P(A\text{ given }B)=P(A)$. That single equation is the test, and it is the only test Cambridge expects.

Two dice are independent. $P(\text{the first dice is a }6)=\frac{1}{6}$, and if you are told the second dice showed a 3, the first is still a 6 with probability $\frac{1}{6}$. The information is irrelevant, so the two numbers match.

Two counters taken from a bag without replacement are not independent. With 3 red and 2 blue, $P(\text{the second is red})=\frac{3}{5}$ - the same as for the first, by symmetry - but $P(\text{the second is red given the first was red})=\frac{2}{4}=\frac{1}{2}$. Since $\frac{1}{2}$ is not $\frac{3}{5}$, the two draws are not independent.

Independence is also what makes the 'and' rule from combined events work. When A and B are independent, $P(B\text{ given }A)=P(B)$, so the multiplication rule becomes $P(A\text{ and }B)=P(A)\times P(B)$. When they are not independent, that shortcut is wrong and the conditional branch has to be used.

• Independent: $P(A\text{ given }B)=P(A)$, and $P(A\text{ and }B)=P(A)\times P(B)$
• Not independent: $P(A\text{ given }B)$ differs from $P(A)$, and the second set of tree branches is unequal
• Telling you the second dice showed a 3 does not change $P(\text{first is a }6)=\frac{1}{6}$, so the dice are independent
• With 3 red and 2 blue taken without replacement, $P(\text{second red})=\frac{3}{5}$ but $P(\text{second red given first red})=\frac{1}{2}$, so the draws are not independent