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Probability - Tree Diagrams

Grade 9IGCSE

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

🔑Concepts

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Tree diagrams are visual tools used to map out all possible outcomes of a sequence of events.

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Branches: Each branch represents a possible outcome. The probability of that outcome is written on the branch.

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The Multiplication Rule (AND): To find the probability of a specific sequence of outcomes (e.g., Outcome A then Outcome B), multiply the probabilities along the chosen branches.

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The Addition Rule (OR): To find the probability of more than one successful combined outcome, add the probabilities of the different paths together.

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Sum of Probabilities: The sum of probabilities on any set of branches originating from the same point must always equal 1.

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Independent vs. Dependent Events: In 'with replacement' scenarios, probabilities remain the same for the second set of branches. In 'without replacement' scenarios, the denominator and/or numerator change for the second set of branches.

📐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 a single node)=1\sum P(\text{branches from a single node}) = 1

P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})

💡Examples

Problem 1:

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random one after the other without replacement. Find the probability that both marbles are the same color.

Solution:

P(RR)+P(BB)=(58×47)+(38×27)=2056+656=2656=1328P(RR) + P(BB) = (\frac{5}{8} \times \frac{4}{7}) + (\frac{3}{8} \times \frac{2}{7}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}

Explanation:

Since the marbles are not replaced, the total number of marbles drops from 8 to 7 for the second draw. The probability of Red then Red (RR) is 5/8 * 4/7. The probability of Blue then Blue (BB) is 3/8 * 2/7. We add these two mutually exclusive paths to get the final result.

Problem 2:

The probability that it rains on Monday is 0.6. If it rains on Monday, the probability it rains on Tuesday is 0.8. If it does not rain on Monday, the probability it rains on Tuesday is 0.3. Calculate the probability that it rains on at least one of the two days.

Solution:

1−P(No rain, No rain)=1−(0.4×0.7)=1−0.28=0.721 - P(\text{No rain, No rain}) = 1 - (0.4 \times 0.7) = 1 - 0.28 = 0.72

Explanation:

To find 'at least one', it is often easier to subtract the probability of 'none' from 1. The probability it doesn't rain on Monday is 0.4 (1 - 0.6). If it doesn't rain Monday, the probability it doesn't rain Tuesday is 0.7 (1 - 0.3). Multiplying these gives the probability of two dry days (0.28).