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Probability - Mutually Exclusive and Independent Events

Grade 12IGCSE

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

🔑Concepts

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Definition of Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur at the same time. If event A happens, event B cannot happen.

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Definition of Independent Events: Two events are independent if the occurrence of one event does not affect the probability of the other event occurring.

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Addition Rule for Mutually Exclusive Events: The probability of A or B occurring is the sum of their individual probabilities because their intersection is zero.

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Multiplication Rule for Independent Events: The probability of both A and B occurring is the product of their individual probabilities.

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Exhaustive Events: A set of events is exhaustive if at least one of them must occur, meaning their total probability sums to 1.

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Intersection and Union: P(A∩B)P(A \cap B) represents both occurring; P(A∪B)P(A \cup B) represents at least one occurring.

📐Formulae

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) (General Addition Rule)

P(A∩B)=0P(A \cap B) = 0 (For Mutually Exclusive Events)

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) (For Mutually Exclusive Events)

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (For Independent Events)

P(A∣B)=P(A)P(A|B) = P(A) (Condition for Independence)

P(A′)=1−P(A)P(A') = 1 - P(A) (Complementary Events)

💡Examples

Problem 1:

Events A and B are such that P(A)=0.3P(A) = 0.3 and P(B)=0.4P(B) = 0.4. If A and B are mutually exclusive, find P(A∪B)P(A \cup B).

Solution:

P(A∪B)=0.3+0.4=0.7P(A \cup B) = 0.3 + 0.4 = 0.7

Explanation:

Since the events are mutually exclusive, P(A∩B)=0P(A \cap B) = 0. We use the simplified addition rule: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B).

Problem 2:

A fair coin is flipped and a fair 6-sided die is rolled. What is the probability of getting a 'Head' and rolling a '6'?

Solution:

P(H∩6)=12×16=112P(H \cap 6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

Explanation:

The outcome of the coin flip does not affect the outcome of the die roll, meaning the events are independent. Therefore, we multiply their individual probabilities.

Problem 3:

Given P(X)=0.5P(X) = 0.5 and P(Y)=0.2P(Y) = 0.2. If X and Y are independent events, find P(X∪Y)P(X \cup Y).

Solution:

P(X∪Y)=0.5+0.2−(0.5×0.2)=0.7−0.1=0.6P(X \cup Y) = 0.5 + 0.2 - (0.5 \times 0.2) = 0.7 - 0.1 = 0.6

Explanation:

First, calculate the intersection using the independence rule P(X∩Y)=P(X)P(Y)=0.1P(X \cap Y) = P(X)P(Y) = 0.1. Then, apply the general addition rule: P(X∪Y)=P(X)+P(Y)−P(X∩Y)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y).