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Probability - Venn Diagrams and Probability

Grade 9IGCSE

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

🔑Concepts

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Universal Set (ξ): The set containing all possible outcomes under consideration, represented by a rectangle.

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Intersection (A ∩ B): The region where sets A and B overlap, representing outcomes that belong to both A AND B.

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Union (A ∪ B): The region covering everything in A, in B, or in both, representing outcomes that belong to A OR B.

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Complement (A'): The region outside set A, representing outcomes that are NOT in A.

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Mutually Exclusive Events: Events that cannot happen at the same time; their circles do not overlap (A ∩ B = ∅).

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Exhaustive Events: Events where the union of the sets covers the entire universal set.

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Probability calculation: The probability of an event is the number of elements in that region divided by the total number of elements in the universal set.

📐Formulae

P(A)=n(A)n(ξ)P(A) = \frac{n(A)}{n(\xi)}

P(A′)=1−P(A)P(A') = 1 - P(A)

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

For mutually exclusive events: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

💡Examples

Problem 1:

In a class of 30 students, 18 play football (F), 15 play basketball (B), and 5 play neither sport. A student is chosen at random. Find the probability that the student plays both sports.

Solution:

P(F∩B)=830=415P(F \cap B) = \frac{8}{30} = \frac{4}{15}

Explanation:

First, find the number of students who play at least one sport: 30−5=2530 - 5 = 25. Let xx be the number of students who play both. Using the formula n(F∪B)=n(F)+n(B)−n(F∩B)n(F \cup B) = n(F) + n(B) - n(F \cap B), we get 25=18+15−x25 = 18 + 15 - x. Solving for xx gives x=33−25=8x = 33 - 25 = 8. The probability is the number of students in the intersection divided by the total class size.

Problem 2:

Given P(A)=0.6P(A) = 0.6, P(B)=0.4P(B) = 0.4, and P(A∪B)=0.8P(A \cup B) = 0.8, calculate P(A∩B′)P(A \cap B').

Solution:

P(A∩B′)=0.4P(A \cap B') = 0.4

Explanation:

First, find the intersection using P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.4−0.8=0.2P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.4 - 0.8 = 0.2. The region (A∩B′)(A \cap B') represents elements in A but NOT in B. This is calculated as P(A)−P(A∩B)=0.6−0.2=0.4P(A) - P(A \cap B) = 0.6 - 0.2 = 0.4.

Problem 3:

A bag contains 20 shapes. 10 are Red (R), 8 are Squares (S), and 3 are Red Squares. Find the probability that a randomly selected shape is neither Red nor a Square.

Solution:

P(R′∩S′)=520=0.25P(R' \cap S') = \frac{5}{20} = 0.25

Explanation:

Calculate the number of shapes that are either Red or Squares: n(R∪S)=n(R)+n(S)−n(R∩S)=10+8−3=15n(R \cup S) = n(R) + n(S) - n(R \cap S) = 10 + 8 - 3 = 15. The number of shapes that are neither is the total minus this union: 20−15=520 - 15 = 5. The probability is 5/205/20.