A probability tree diagram represents a multi-stage experiment: from each node, branches lead to every possible outcome at the next stage, and each branch is labelled with its probability. A complete path from root to leaf describes one full scenario.

The two golden rules:

• Multiply along a path: the probability of a complete scenario = product of branch probabilities along that path (intersection of events).
• Add across paths: when several different paths all satisfy the condition, add their probabilities (union of mutually exclusive events).

With replacement vs. without replacement:

• With replacement — the drawn item is returned, so the composition is unchanged. Probabilities at every stage are identical and the stages are independent.
• Without replacement — the item is not returned, so the composition and total change, and branch probabilities at each subsequent stage update (conditional probability).

A useful shortcut is the complementary event: \( P(\text{at least one}) = 1 - P(\text{none}) \). Computing 'none' usually requires only a single path.

Example 1: With Replacement — Two Identical Draws

Problem: A jar contains \(4\) red balls and \(6\) blue balls. A ball is drawn, replaced, and a second ball is drawn. What is the probability that both balls are red?

Solution:

1. There are \(10\) balls in total; since the ball is replaced, the composition is unchanged and the two draws are independent.
2. Probability of red on each draw: \( \frac{4}{10} = 0.4 \).
3. Multiply along the path 'red \(\to\) red': \( 0.4 \times 0.4 = 0.16 \).
4. Equivalently: \( \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} \).

Answer: The probability is \( 0.16 \).

Example 2: Without Replacement — Same Question

Problem: From the same jar (\(4\) red, \(6\) blue) two balls are drawn without replacement. What is the probability that both are red?

Solution:

1. First draw: \( \frac{4}{10} \).
2. After removing a red ball, \(3\) red balls remain among \(9\) total, so the second draw: \( \frac{3}{9} \).
3. Multiply along the path: \( \frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \).
4. As a decimal: \( \frac{2}{15} \approx 0.133 \) — lower than \(0.16\) from the with-replacement case, as expected.

Answer: The probability is \( \frac{2}{15} \approx 0.133 \).

Example 3: Adding Paths — Exactly One Red Ball

Problem: From the jar (\(4\) red, \(6\) blue), two balls are drawn with replacement. What is the probability of getting exactly one red ball?

Solution:

1. 'Exactly one red' occurs via two mutually exclusive paths: red–blue, or blue–red.
2. Path red \(\to\) blue: \( \frac{4}{10} \cdot \frac{6}{10} = \frac{24}{100} \).
3. Path blue \(\to\) red: \( \frac{6}{10} \cdot \frac{4}{10} = \frac{24}{100} \).
4. Add across paths: \( \frac{24}{100} + \frac{24}{100} = \frac{48}{100} = 0.48 \).

Answer: The probability is \( 0.48 \).

Example 4: Complementary Event — At Least One

Problem: A fair die is rolled three times. What is the probability of obtaining the number \(6\) at least once?

Solution:

1. Direct calculation would require summing many paths, so we use the complement: 'never a \(6\)'.
2. Probability of not rolling \(6\) on one roll: \( \frac{5}{6} \); the rolls are independent.
3. Never a \(6\) in three rolls: \( \left(\frac{5}{6}\right)^3 = \frac{125}{216} \).
4. Therefore \( P(\text{at least one}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421 \).

Answer: The probability is \( \frac{91}{216} \approx 0.421 \).

Example 5: Three-Stage Path Without Replacement

Problem: A box contains \(5\) winning and \(5\) losing tickets. Three tickets are drawn one after another without replacement. What is the probability that all three are winning?

Solution:

1. There are \(10\) tickets in total. Both numerator and denominator update at each stage.
2. Stage 1: \( \frac{5}{10} \); stage 2 (\(4\) winning among \(9\) remaining): \( \frac{4}{9} \); stage 3 (\(3\) among \(8\)): \( \frac{3}{8} \).
3. Multiply along the path: \( \frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} = \frac{60}{720} \).
4. Simplify: \( \frac{60}{720} = \frac{1}{12} \approx 0.083 \).

Answer: The probability is \( \frac{1}{12} \approx 0.083 \).

✗ Common mistake: Using the same probabilities at every stage even when sampling is without replacement.

✓ The correct way: Without replacement, the composition changes after each draw. Update both the numerator (how many of that type remain) and the denominator (total items remaining) on every branch.

✗ Common mistake: Adding probabilities along a single path instead of multiplying them.

✓ The correct way: Along a path (stage after stage) you multiply, because it is an intersection of events. Adding is reserved for separate paths that all satisfy the same condition.

✗ Common mistake: For 'exactly one' questions, counting only one path and forgetting the reverse order.

✓ The correct way: Include all orderings. 'Exactly one red' covers both red–blue and blue–red, so both paths must be added.

• Along a path: multiply the branch probabilities.
• Across paths: add the probabilities of mutually exclusive paths.
• With replacement: fixed probabilities, independent stages.
• Without replacement: probabilities update (conditional).
• At least one: \( P = 1 - P(\text{none}) \).
• Branch probabilities at every node sum to \( 1 \).