Using Tree Diagrams

What Is a Tree Diagram?

A tree diagram is a handy way to show all the possible outcomes of one or more events, especially when each event has two or more outcomes. It’s called a “tree” because it branches out like one!

You read a tree diagram from left to right. Each branch shows a possible outcome and its probability.

How to Draw One

1. Start with the first event – draw a branch for each possible outcome.
2. From the end of each branch, draw the branches for the second event.
3. Write the probabilities on each branch.
4. Make sure the branches from the same point always add up to 1.

How to Use a Tree Diagram

Multiply along the branches

This gives you the probability of both events happening:

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

Add between branches

When you want either event:

$$P(A \text{ or } B) = P(A) + P(B)$$

Use subtraction for “at least one”

$$P(\text{At least one A}) = 1 - P(\text{No A})$$

Example 1: Spinning and Tossing

A spinner has two options: Red (R) and Blue (B).

A coin is flipped afterward: Heads (H) or Tails (T).

• $P(R) = 0.6 \to \text{ So, } P(B) = 1 - 0.6 = 0.4$
• $P(H) = 0.5 \to P(T) = 0.5$

Question: What’s the probability of getting a Blue and Heads?

$$P(B, H) = 0.4 \times 0.5 = 0.2$$

Example 2: Picking Sweets Without Replacement

A bag has 5 strawberry and 3 lemon sweets.

Sarah picks one sweet at random, eats it, and then picks another.

Step 1: First pick

• P(Strawberry first) = $\frac{5}{8}$
• P(Lemon first) = $\frac{3}{8}$

Step 2: Second pick (depends on what was picked first)

• If Strawberry first $\to$ 4 strawberry and 3 lemon left
  • P(Strawberry second) = $\frac{4}{7}$
  • P(Lemon second) = $\frac{3}{7}$
• If Lemon first $\to$ 5 strawberry and 2 lemon left
  • P(Strawberry second) = $\frac{5}{7}$
  • P(Lemon second) = $\frac{2}{7}$

Question: What’s the probability Sarah picks one of each?

That’s two cases:

• Strawberry then Lemon: $\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$
• Lemon then Strawberry: $\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$

Add both:

$$\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$$

Example: Conditional Probabilities

A basketball player has a 60% chance of scoring a free throw.

If they score the first shot, their confidence rises, and their chance of scoring the second goes up to 80%.

If they miss the first, the second shot success drops to 40%.

Question: What is the probability of scoring both shots?

$$P(\text{Score, Score}) = 0.6 \times 0.8 = 0.48$$