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

How to use

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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

0 / 3 correct0 / 3 correct

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?