Learning Resources

Tree Diagrams & Combined Events

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

N/A Pathway: N/A

First Principles

Objective: Master tree diagrams to calculate probabilities of combined independent and dependent events (with and without replacement).

The Big Idea: Probability problems with multiple stages can quickly become confusing. A tree diagram maps out every possible sequence of events step-by-step. Each "branch" represents a choice or outcome, and each "path" from root to leaf represents one complete combined outcome.

Core Rules of Tree Diagrams

Rule 1 (Along a Branch): Multiply probabilities along a path to find the probability of that specific combined outcome: \(P(A \text{ and } B) = P(A) \times P(B \mid A)\).
Rule 2 (Across Outcomes): Add the probabilities of all distinct paths that fulfill the required condition: \(P(\text{Event}) = \sum P(\text{Favourable Paths})\).
Rule 3 (Branch Totals): The sum of probabilities branching out from any single node must always equal 1.

With vs. Without Replacement:

With replacement (Independent): Items are returned. Probabilities stay constant on the 2nd draw (denominator does not change).
Without replacement (Dependent): Items are kept. Total items decrease by 1, and the count of the chosen colour decreases by 1 (denominator shrinks to \(n-1\)).

Interactive Tree Diagram Simulator

Simulate picking 2 beads from a pouch with 3 Red and 2 Blue beads. Toggle replacement mode to see how probabilities change!

Key Formulas

1. Independent Events (With Replacement) \[P(A \cap B) = P(A) \times P(B)\] The outcome of the first event has no effect on the probability of the second.

2. Dependent Events (Without Replacement) \[P(A \cap B) = P(A) \times P(B \mid A)\] Where \(P(B \mid A)\) is the conditional probability of event \(B\) occurring given that event \(A\) has already happened.

3. Product Rule (Along a Path) \[P(\text{Path}) = P(\text{Branch}_1) \times P(\text{Branch}_2) \times \dots \times P(\text{Branch}_n)\]

4. Addition Rule (Combining Mutually Exclusive Paths) \[P(\text{Event}) = \sum_{\text{matching paths}} P(\text{Path})\]

5. Complement Rule ("At Least One") \[P(\text{at least one } A) = 1 - P(\text{no } A)\] Crucial time-saver for multi-stage problems.

Worked Examples

Example 1 (Easy): Two draws without replacement — matching pair

A pouch contains 4 green and 2 yellow marbles. Two marbles are drawn one after another without replacement. Find the probability that both are green.

Step 1: Identify the probability of the first marble being green.

\[P(G_1) = \frac{4}{6} = \frac{2}{3}\]

Step 2: Calculate the conditional probability of the second marble being green (1 green removed, total decreases by 1).

\[P(G_2 \mid G_1) = \frac{3}{5}\]

Step 3: Multiply along the path:

\[P(G_1 \cap G_2) = \frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5}\]

Example 2 (Medium): Mixed outcomes — "One of each colour"

A vendor in Kisumu has a crate with 7 sweet mangoes and 3 sour mangoes. A buyer picks two mangoes at random without replacement. What is the probability that the buyer gets one sweet and one sour mango?

Step 1: Identify all possible paths that satisfy the condition.

Path 1: Sweet then Sour \((S_1, R_2)\)
Path 2: Sour then Sweet \((R_1, S_2)\)

Step 2: Calculate the probability for each path.

\[P(S_1 \cap R_2) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90}\] \[P(R_1 \cap S_2) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90}\]

Step 3: Add the mutually exclusive paths together:

\[P(\text{One of each}) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90} = \frac{7}{15}\]

Example 3 (Hard): Multi-stage with Complement ("At Least One")

A bag contains 5 functional USB drives and 3 corrupted USB drives. A technician selects 3 drives at random without replacement. Find the probability that at least one functional drive is selected.

Strategy: Using the direct method requires summing \(P(1), P(2),\) and \(P(3)\) functional drives (7 different paths!). Instead, use the complement rule: \(P(\text{at least 1 functional}) = 1 - P(\text{all 3 corrupted})\).

Step 1: Calculate \(P(\text{all corrupted}) = P(C_1 \cap C_2 \cap C_3)\):

\[P(C_1 \cap C_2 \cap C_3) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{6}{336} = \frac{1}{56}\]

Step 2: Subtract from 1:

\[P(\text{at least one functional}) = 1 - \frac{1}{56} = \frac{55}{56}\]

Common Mistakes

Mistake 1: Keeping denominators unchanged in "without replacement" problems

Incorrect: Calculating \(P(R_1 \cap R_2) = \frac{3}{5} \times \frac{2}{5}\).

Correction The total number of items in the container decreases after each draw! Denominator drops from \(5\) to \(4\): \(P(R_1 \cap R_2) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}\).

Why it feels right Your mental image of the bag remains static unless you explicitly visualize physically removing an item.

Mistake 2: Forgetting order in "one of each" questions

Incorrect: Answering \(P(\text{Red and Blue}) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}\).

Correction "One red and one blue" means Red-then-Blue OR Blue-then-Red. You must compute and add both paths: \(\frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}\).

Mistake 3: Adding probabilities along a path instead of multiplying

Incorrect: \(P(A \text{ then } B) = P(A) + P(B)\).

Correction Combined sequential events obey the Product Rule: Multiply along branches (\(\times\)). You only add (\(+\)) when combining separate parallel paths.

Real World

Matatu Commuter Reliability: In Nairobi, there is a \(0.35\) probability of heavy morning rain. If it rains, the probability of traffic delay is \(0.80\); if dry, delay probability is \(0.20\). A tree diagram models the daily commute: \[P(\text{Delay}) = (0.35 \times 0.80) + (0.65 \times 0.20) = 0.28 + 0.13 = 0.41 \text{ (41\%)}\]

Mobile Money Fraud Detection: An automated fraud detection filter flags \(95\%\) of fraudulent transactions and falsely flags \(2\%\) of legitimate transactions. If \(1\%\) of all transactions are fraudulent, tree diagrams calculate the exact probability that a flagged transaction is genuinely fraudulent.

Medical Testing & Diagnostics: Rapid antigen testing for malaria uses conditional tree branches to evaluate true positive and false positive rates across patient populations.

Practice

A fair coin is tossed and a fair six-sided die is rolled. What is the probability of obtaining a Head on the coin and an even number on the die? (Type only the fraction in simplest form, e.g., 1/4)

Review the concepts above.

A bag contains 3 red balls and 2 blue balls. Two balls are drawn successively **with replacement**. What is the probability that both balls drawn are red? (Type only the fraction in simplest form, e.g., 9/25)

Review the concepts above.

A bag contains 3 red, 2 blue, and 1 green ball (total 6 balls). Two balls are drawn successively **without replacement**. What is the probability that both balls drawn are red? (Type only the fraction in simplest form, e.g., 1/5)

Review the concepts above.

A factory produces items using two machines. 60% of the items come from Machine A and 40% from Machine B. The probability that an item from Machine A is defective is 0.02 (2%), and from Machine B is 0.05 (5%). What is the overall probability that a randomly selected item is defective? (Type your answer as a simplified fraction, e.g., 4/125)

Review the concepts above.

A basket holds 4 ripe mangoes and 3 unripe mangoes. A shopper picks 2 mangoes at random **without replacement**. What is the probability that **at least one** mango is ripe? (Type only the fraction in simplest form, e.g., 6/7)

Review the concepts above.

A quality control line inspects items. An item is Type A with probability 0.6 and Type B with probability 0.4. Type A items pass inspection with probability 0.9, while Type B items pass with probability 0.7. What is the probability that a randomly selected item is either Type B OR passes inspection? (Type only the decimal, e.g., 0.94)

Review the concepts above.