A sample space diagram lists every possible combined outcome of two events in a grid, making favourable outcomes easy to count.

Two fair 4-sided spinners are spun and their scores multiplied:

Find $P(\text{product is a multiple of }4)$: total outcomes $=16$; counting the multiples of $4$ in the grid gives $10$ cells, so

A Venn diagram (limited to two sets at Core level) shows how many outcomes belong to each event, to both (the overlap), or to neither, using $n(A)$, $A\cup B$ and $A\cap B$.

In a class of $30$, $18$ study French, $14$ study Spanish, and $7$ study both. Find $P(\text{neither})$:

1. Both: $n(F\cap S)=7$ — the overlap.
2. French only: $18-7=11$. Spanish only: $14-7=7$.
3. Neither: $30-(11+7+7)=5$.

Once the diagram is filled in correctly, other probabilities become easy to read straight off it too — for instance, $P(\text{French only}) = \dfrac{11}{30}$, and $P(\text{at least one language}) = \dfrac{11+7+7}{30} = \dfrac{25}{30} = \dfrac{5}{6}$.

A tree diagram shows sequential events along branches, and at Core level combined events are always with replacement — each draw doesn't affect the next.

• Multiply along a single branch path for the probability of that whole sequence: $P(A\text{ and }B)=P(A)\times P(B)$.
• Add different paths together when more than one route leads to the outcome you want.

A bag has $3$ red and $2$ blue counters. A counter is drawn, replaced, then a second is drawn. Find $P(\text{both same colour})$:

1. Since replaced, $P(\text{red})=\tfrac35$ and $P(\text{blue})=\tfrac25$ on both draws.
2. "Red then Red": $\dfrac35\times\dfrac35=\dfrac{9}{25}$.
3. "Blue then Blue": $\dfrac25\times\dfrac25=\dfrac{4}{25}$.
4. Add the two paths: $\dfrac{9}{25}+\dfrac{4}{25}=\dfrac{13}{25}$.

A bag contains $4$ green and $6$ orange counters. A counter is drawn, replaced, then a second is drawn. Find the probability that the two counters are different colours.

Solution

1. $P(\text{green})=\tfrac{4}{10}=\tfrac25$, $P(\text{orange})=\tfrac{6}{10}=\tfrac35$.
2. "Green then Orange": $\dfrac25\times\dfrac35=\dfrac{6}{25}$.
3. "Orange then Green": $\dfrac35\times\dfrac25=\dfrac{6}{25}$.
4. Add the two paths (different orders both count):