Probability Trees

Probability trees visualize multi-stage experiments with conditional probabilities.

Building a Tree

1. Draw branches for each outcome at stage 1 (with probabilities)
2. From each branch, draw branches for stage 2 (with conditional probabilities)
3. Continue for all stages

Multiplication Rule

Multiply probabilities along branches to find the probability of any path through the tree:

$$P(A \cap B) = P(A) \cdot P(B | A)$$

Reading the formula:

• $P(A)$ — probability of the first event
• $P(B | A)$ — probability of the second event given the first one already happened
• The product gives the probability that both events occur in sequence

Why it works: The definition of conditional probability says $P(B | A) = \frac{P(A \cap B)}{P(A)}$. Multiply both sides by $P(A)$ and you get the multiplication rule. On a tree, you’re simply multiplying the probabilities along a path from root to leaf.

The key question: does the first event change the second?

The multiplication rule hinges on $P(B \vert A)$ — the probability of the second event after the first has occurred. But that raises a natural question: does the first event actually change the odds for the second one, or not?

Consider drawing marbles from a bag. After you draw the first marble, you have a choice: put it back or keep it out. This choice determines whether the first draw changes the situation for the second.

• With replacement — you put the marble back. The bag is exactly the same as before, so the second draw is unaffected by the first.
• Without replacement — you keep the marble out. The bag now has one fewer marble, so the odds shift.

This matters for the multiplication rule because $P(B \vert A)$ — the conditional probability in the formula — depends on whether the first event changed the situation:

Example: A bag has 3 red and 2 blue marbles (5 total). Draw two marbles.

With replacement (put the first marble back):

• $P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red} | \text{1st red}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} = 0.36$
• After replacing, the bag is the same — so $P(\text{2nd red} | \text{1st red}) = P(\text{2nd red}) = \frac{3}{5}$

Without replacement (keep the first marble):

• $P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red} | \text{1st red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = 0.3$
• After removing a red, only 2 red remain out of 4 — the conditional probability drops to $\frac{2}{4}$

The multiplication rule is the same formula either way. The difference is what number goes into $P(B | A)$ — and that depends on whether the first draw changes the situation.

General Chain Rule

For three or more events, the rule extends naturally:

$$P(A_1 \cap A_2 \cap A_3) = P(A_1) \cdot P(A_2 | A_1) \cdot P(A_3 | A_1 \cap A_2)$$

Reading the formula:

• $P(A_1)$ — probability of the first event
• $P(A_2 | A_1)$ — probability of the second event given the first
• $P(A_3 | A_1 \cap A_2)$ — probability of the third event given both the first and second

Each factor conditions on all previous events — not just the immediately preceding one. This matters: $P(A_3 | A_1 \cap A_2)$ is not the same as $P(A_3 | A_2)$ when $A_1$ carries additional information.

The pattern continues for any number of events. For four events:

$$P(A_1 \cap A_2 \cap A_3 \cap A_4) = P(A_1) \cdot P(A_2 | A_1) \cdot P(A_3 | A_1 \cap A_2) \cdot P(A_4 | A_1 \cap A_2 \cap A_3)$$

Just keep adding one conditional factor at a time — each new factor conditions on everything that came before.

Example: Draw three cards from a deck without replacement. Find $P(\text{three Aces})$.

• $P(\text{1st Ace}) = \frac{4}{52}$
• $P(\text{2nd Ace} | \text{1st Ace}) = \frac{3}{51}$ — one Ace removed, 51 cards left
• $P(\text{3rd Ace} | \text{1st two Aces}) = \frac{2}{50}$ — two Aces removed, 50 cards left
• $P(\text{three Aces}) = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \approx 0.00018$

Repeated Trials

What does the tree mean if we run the same experiment many times?

• $P(A_1)$ — out of all trials, the proportion where $A_1$ happens
• $P(A_2 | A_1)$ — out of only the trials where $A_1$ happened, the proportion where $A_2$ also happens
• Multiplying them gives the proportion of all trials where both $A_1$ and $A_2$ happen

The tree captures this step-by-step filtering process visually — each branch narrows down the set of trials, and multiplying along a path gives the proportion that makes it all the way to that leaf.

Example: Two Coin Flips

Each leaf has probability $0.5 \times 0.5 = 0.25$.

Example: Disease Testing

A test for a disease (1% prevalence):

• Test positive if diseased: 95%
• Test positive if healthy: 10%

Summing Probabilities

To find P(positive), sum all paths ending in positive:

$$P(+) = 0.0095 + 0.099 = 0.1085$$