Statistics · Conditional Probability · Year 2

Conditional Probability in Venn Diagrams

Shade the region you're given, then ask what fraction of it you actually want.

7 min read 1 worked example

Restricting the sample space visually

You can find conditional probabilities straight from a Venn diagram, by considering the section of the diagram that corresponds to the restricted sample space — the "given" event becomes the new whole, and you ask what fraction of it satisfies the rest of the condition.

Worked example

Reading three conditional probabilities off one Venn diagram

$A$ and $B$ are two events such that $\mathrm{P}(A)=0.45$, $\mathrm{P}(B)=0.35$ and $\mathrm{P}(A\cap B)=0.1$.

a Draw a Venn diagram showing the probabilities for events $A$ and $B$.

The probabilities for each of the four regions, filled in from $\mathrm{P}(A)=0.45$, $\mathrm{P}(B)=0.35$ and $\mathrm{P}(A\cap B)=0.1$.

MethodFill in the intersection first, then work outwards

The intersection is given directly: $\mathrm{P}(A\cap B)=0.1$.

$A$ only $= \mathrm{P}(A)-\mathrm{P}(A\cap B) = 0.45-0.1=0.35$.

Subtract the overlap from the whole of A.

$B$ only $= \mathrm{P}(B)-\mathrm{P}(A\cap B) = 0.35-0.1=0.25$.

Subtract the overlap from the whole of B.

Outside both $= 1-0.35-0.1-0.25=0.3$.

The four regions must sum to 1.

▪ solved

b Find: i $\mathrm{P}(A\mid B)$ ii $\mathrm{P}(B\mid A\cup B)$ iii $\mathrm{P}(A'\mid B')$

i — P(A|B)

To find $\mathrm{P}(A\mid B)$, restrict the sample space to just circle $B$ (shaded) — both regions inside it.

$\mathrm{P}(A\mid B) = \dfrac{0.1}{0.1+0.25} = \dfrac{0.1}{0.35} = \dfrac{2}{7}$.

The sample space is restricted to just circle $B$. The part of circle $A$ inside $B$ has probability $0.1$; circle $B$ in total has probability $0.35$.

ii — P(B | A∪B)

To find $\mathrm{P}(B\mid A\cup B)$, restrict the sample space to the union of $A$ and $B$ (shaded).

$\mathrm{P}(B\mid A\cup B) = \dfrac{0.1+0.25}{0.35+0.1+0.25} = \dfrac{0.35}{0.7} = \dfrac{1}{2}$.

The sample space is restricted to just the union of $A$ and $B$; within it, every region that's part of $B$ is in the numerator.

iii — P(A' | B')

To find $\mathrm{P}(A'\mid B')$, restrict the sample space to everything outside circle $B$ (shaded) — this is $B'$.

$\mathrm{P}(A'\mid B') = \dfrac{0.3}{0.35+0.3} = \dfrac{0.3}{0.65} = \dfrac{6}{13}$.

Consider the restricted sample space first — this is everything not inside circle $B$. Within it, the numerator is everything also not inside $A$.

The pattern behind all three parts

Every conditional probability here follows the same two steps: shade the region given by the condition (after the "$\mid$"), then find what fraction of that shaded region also satisfies the rest of the statement. Getting the shading right is most of the work — the arithmetic that follows is just addition and division.

Common mistake

For $\mathrm{P}(A'\mid B')$, it's tempting to restrict to circle $A'$ instead of $B'$ — but the event after the "$\mid$" is always the one that defines the restricted sample space, regardless of which event is named first in the statement.

Check yourself

Using the same Venn diagram ($\mathrm{P}(A)=0.45$, $\mathrm{P}(B)=0.35$, $\mathrm{P}(A\cap B)=0.1$), find $\mathrm{P}(B\mid A)$.

Restrict to circle $A$ (probability $0.35+0.1=0.45$, as expected).

$\mathrm{P}(B\mid A) = \dfrac{0.1}{0.45} = \dfrac{2}{9}$.

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