Independent: the first outcome doesn't affect the second — e.g. flipping a coin twice, or picking a counter and replacing it before picking again.

Dependent: the first outcome changes the probabilities for the second — most commonly, picking a counter and not replacing it, so both the total and the count of that colour drop by one.

1. Draw one branch per outcome from each node, labelled with its probability.
2. Check: probabilities from any single node must sum to exactly 1.
3. For dependent events, recalculate the second-stage branches based on what happened first ("Think, Adjust, Calculate": what happened → what's the new total → what's the new probability).
4. Multiply along each full path (the AND rule) to get that path's final probability.
5. Final check: every final-outcome probability should sum to 1.

Multiply along a path (AND), add across different qualifying paths (OR):

Example (independent, with replacement): A bag has 5 red, 3 blue counters; one is picked, replaced, then a second is picked. $P(\text{different colours}) = P(R,B)+P(B,R) = \tfrac58\times\tfrac38 + \tfrac38\times\tfrac58 = \tfrac{15}{64}+\tfrac{15}{64}=\tfrac{30}{64}=\tfrac{15}{32}$.

The total drops by one, and so does the count for whichever colour was picked first.

Example: 7 blue, 5 green counters (12 total), picked without replacement.

• First pick: $P(\text{Blue})=\tfrac{7}{12}$, $P(\text{Green})=\tfrac{5}{12}$.
• If first was Blue, remaining: 6 blue, 5 green out of 11 → $P(\text{Blue}|\text{Blue})=\tfrac6{11}$.
• If first was Green, remaining: 7 blue, 4 green out of 11 → $P(\text{Blue}|\text{Green})=\tfrac7{11}$.

Worked Example

A bag contains 7 blue and 5 green counters. A counter is taken, not replaced, then a second is taken. Find $P(\text{same colour})$.

1. $P(B,B)=\tfrac{7}{12}\times\tfrac6{11}=\tfrac{42}{132}$
2. $P(G,G)=\tfrac5{12}\times\tfrac4{11}=\tfrac{20}{132}$
3. Check: all four path probabilities sum to $\tfrac{132}{132}=1$. ✓
4. $P(\text{same}) = \tfrac{42}{132}+\tfrac{20}{132}=\tfrac{62}{132}=\tfrac{31}{66}$

Probability tree questions are almost always split into two parts: (a) complete the diagram, (b) work out a probability. Treat these as two separate skills. Secure part (a) first by being meticulous with every branch label and probability — a correct, fully-labelled diagram earns marks in its own right, independently of whether part (b) is answered correctly afterwards. Only once the diagram is solid should you move on to selecting and combining the right paths for part (b). This staged approach means a slip in the final calculation doesn't cost you the marks already secured for a correct diagram.