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Probability - Probability Tree Diagrams

Grade 12IGCSE

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

🔑Concepts

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Tree diagrams are visual representations used to list all possible outcomes of a sequence of events.

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Each branch represents a possible outcome, and the probability of that outcome is written on the branch.

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The sum of the probabilities on any set of branches originating from a single point must always equal 1.

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Multiplication Rule: To find the probability of a specific combination of outcomes, multiply the probabilities along the branches of that path (the 'AND' rule).

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Addition Rule: To find the probability of more than one combined outcome (e.g., 'at least one'), add the probabilities of the relevant end-nodes (the 'OR' rule).

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Independent Events: The outcome of the first event does not affect the second; probabilities remain constant on subsequent branches.

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Dependent Events (Without Replacement): The outcome of the first event changes the probabilities of the second event, common in 'picking from a bag' scenarios without replacement.

📐Formulae

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

P(Event)=∑(Probabilities of all successful paths)P(\text{Event}) = \sum (\text{Probabilities of all successful paths})

∑P(outcomes at any node)=1\sum P(\text{outcomes at any 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. Calculate the probability that both marbles are the same color.

Solution:

P(Same Color)=P(RR)+P(BB)=(58×47)+(38×27)=2056+656=2656=1328P(\text{Same Color}) = P(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 decreases from 8 to 7 for the second draw. For the 'Red then Red' path, the first probability is 5/8 and the second is 4/7. For the 'Blue then Blue' path, the first is 3/8 and the second is 2/7. We multiply along the branches and then add the results of the two valid paths.

Problem 2:

The probability that it rains on any given day is 0.3. If it rains, the probability that a student is late for school is 0.4. If it does not rain, the probability that the student is late is 0.1. Find the probability that the student is late on a randomly selected day.

Solution:

P(Late)=P(Rain∩Late)+P(No Rain∩Late)=(0.3×0.4)+(0.7×0.1)=0.12+0.07=0.19P(\text{Late}) = P(\text{Rain} \cap \text{Late}) + P(\text{No Rain} \cap \text{Late}) = (0.3 \times 0.4) + (0.7 \times 0.1) = 0.12 + 0.07 = 0.19

Explanation:

This is a conditional probability problem represented by a tree diagram. The first set of branches is Rain (0.3) and No Rain (0.7). From Rain, the sub-branches are Late (0.4) and Not Late (0.6). From No Rain, the sub-branches are Late (0.1) and Not Late (0.9). To find the total probability of being late, we add the results of the two paths that end in 'Late'.