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Probability - Tree diagrams

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition: A visual tool used to represent all possible outcomes of a sequence of events.

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Nodes and Branches: Each node represents a point of decision or outcome; branches represent the possible choices/results from that point.

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Sum of Probabilities: The probabilities on the branches originating from any single node must always sum to 1.

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Multiplication Rule (Along Branches): To find the probability of a specific sequence of outcomes, multiply the probabilities along the path (representing the 'AND' rule).

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Addition Rule (Multiple Outcomes): To find the probability of more than one combined outcome, add the probabilities of the relevant end-points of the tree (representing the 'OR' rule).

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Independent Events (With Replacement): The probabilities remain the same for the second set of branches.

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Dependent Events (Without Replacement): The probabilities for the second set of branches change based on the outcome of the first event (e.g., the denominator decreases).

📐Formulae

P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B|A)

P(Outcome 1 or Outcome 2)=P(Path 1)+P(Path 2)P(\text{Outcome 1 or Outcome 2}) = P(\text{Path 1}) + P(\text{Path 2})

∑P(branches from one node)=1\sum P(\text{branches from one node}) = 1

P(At least one)=1−P(None)P(\text{At least one}) = 1 - P(\text{None})

💡Examples

Problem 1:

A bag contains 7 blue marbles and 3 red marbles. Two marbles are drawn one after the other without replacement. Draw a tree diagram and find the probability that both marbles are the same color.

Solution:

  1. First draw: P(B)=7/10P(B) = 7/10, P(R)=3/10P(R) = 3/10.
  2. Second draw (if first was Blue): P(B)=6/9P(B) = 6/9, P(R)=3/9P(R) = 3/9.
  3. Second draw (if first was Red): P(B)=7/9P(B) = 7/9, P(R)=2/9P(R) = 2/9.
  4. Same color paths: P(B,B)=(7/10×6/9)=42/90P(B,B) = (7/10 \times 6/9) = 42/90; P(R,R)=(3/10×2/9)=6/90P(R,R) = (3/10 \times 2/9) = 6/90.
  5. Total: 42/90+6/90=48/90=8/1542/90 + 6/90 = 48/90 = 8/15.

Explanation:

Because the marbles are not replaced, the total number of marbles decreases to 9 for the second draw, and the count of the color picked first also decreases by 1. We identify the two 'same color' paths (Blue-Blue and Red-Red), multiply along their branches, and add the results.

Problem 2:

The probability that it rains on any given day is 0.4. If it rains, the probability that John takes the bus is 0.8. If it does not rain, the probability he takes the bus is 0.3. Find the probability that it rains and John does NOT take the bus.

Solution:

P(Rain)=0.4P(\text{Rain}) = 0.4, so P(No Rain)=0.6P(\text{No Rain}) = 0.6. Given Rain: P(Bus)=0.8P(\text{Bus}) = 0.8, so P(No Bus)=0.2P(\text{No Bus}) = 0.2. Calculation: P(Rain and No Bus)=0.4×0.2=0.08P(\text{Rain and No Bus}) = 0.4 \times 0.2 = 0.08.

Explanation:

This is a conditional probability problem. We follow the 'Rain' branch and then the 'No Bus' branch. By multiplying the probabilities along that specific path, we find the intersection of both events.