Compound Probability & Tree Diagrams

In the introduction to probability you learned one-step chances: a coin lands heads with probability $\frac{1}{2}$ . But life often runs two steps in a row — flipping two coins, drawing two cards, rolling a die and then flipping a coin. How do two-step probabilities work? The answer starts with drawing a tree.

The tree diagram: drawing the forks

Flip two coins. Each coin is a fork in the road. Draw it:

First coin heads
- Second coin heads → heads-heads
- Second coin tails → heads-tails
First coin tails
- Second coin heads → tails-heads
- Second coin tails → tails-tails

There are $2 \times 2 = 4$ complete paths from root to tip. Each path passes through two $\frac{1}{2}$ choices, so every path has probability

$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Four equally likely paths, a quarter each — once the tree is drawn, the answers are already sitting on the paper.

Multiply along branches, add across forks

Two slogans hide behind the tree:

"And" means multiply: when the desired result must pass several gates in a row, multiply the gate probabilities. Two heads? $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ;
"Or" means add: when several non-overlapping paths all count, add their probabilities. "At least one head" covers heads-heads, heads-tails and tails-heads: $\frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$ .

'At least one' is bigger than intuition

Many people guess one half for "at least one head", because it sounds like a single event. On the tree it occupies three paths — only tails-tails fails. Counting non-overlapping paths one by one is the greatest service the tree diagram performs.

Die plus coin

Roll a die and flip a coin; find "a six and heads". The two steps do not disturb each other (independent events), so multiply along the branch: $P = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$ . What about "a six or heads"? $\frac{1}{6} + \frac{1}{2} - \frac{1}{12} = \frac{7}{12}$ — the path satisfying both, $\frac{1}{12}$ , may only be added once.

With replacement, or without

A bag holds 3 marbles, 2 red and 1 blue; draw twice. If each marble goes back in, the second gate keeps its probability: two reds is $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$ . Without replacement, after drawing a red the bag holds 1 red and 1 blue, so the second gate changes: two reds is $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ . The tree still works — the second layer of branches just carries new probabilities. Before drawing any tree, always ask: with replacement or without?

InteractiveIndependent Events Lab

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Heads rate

The more you flip, the closer the heads rate gets to 50% — that's the law of large numbers.

Multiplying along branches is only valid when the two steps do not interfere. Flip 100, then 1000 times and watch: the heads ratio settles tightly around $\frac{1}{2}$ — every flip starts from scratch, touching nothing else. That is what "independent" means.

Check yourself

Quick quiz

1. Two coins are flipped. What is the probability of exactly one head and one tail?
2. A die is rolled and a coin flipped. What is the probability of 'a six and heads'?
3. A bag holds 2 red and 1 blue marble. Without replacement, what is the chance both draws are red?