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Probability - Venn Diagrams in Probability

Grade 10IGCSE

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

πŸ”‘Concepts

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Universal Set (ΞΎ): The set containing all possible outcomes in the probability experiment, represented by a rectangle.

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

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Union (A βˆͺ B): The region covering all areas within set A or set B or both, representing '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 in a Venn diagram do not overlap, so P(A ∩ B) = 0.

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Exhaustive Events: Events where the union of their sets covers the entire universal set, meaning P(A βˆͺ B) = 1 (if no outcomes lie outside the circles).

πŸ“Formulae

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

P(AβˆͺB)=P(A)+P(B)βˆ’P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) (The General Addition Rule)

P(Aβ€²)=1βˆ’P(A)P(A') = 1 - P(A)

P(A∩Bβ€²)=P(A)βˆ’P(A∩B)P(A \cap B') = P(A) - P(A \cap B) (Probability of 'A only')

P(A∣B)=n(A∩B)n(B)P(A | B) = \frac{n(A \cap B)}{n(B)} (Conditional probability using frequencies)

πŸ’‘Examples

Problem 1:

In a class of 30 students, 18 study Biology (B), 15 study Chemistry (C), and 7 study both. A student is chosen at random. Find the probability that the student studies: i) Only Biology, ii) Neither Biology nor Chemistry.

Solution:

i) P(OnlyΒ B)=18βˆ’730=1130P(\text{Only B}) = \frac{18 - 7}{30} = \frac{11}{30}. ii) Total in circles = (18βˆ’7)+7+(15βˆ’7)=11+7+8=26(18-7) + 7 + (15-7) = 11 + 7 + 8 = 26. Students outside = 30βˆ’26=430 - 26 = 4. P(BβˆͺC)β€²=430=215P(B \cup C)' = \frac{4}{30} = \frac{2}{15}.

Explanation:

First, fill the intersection (7). Subtract 7 from the totals of B and C to find those who study 'only' one subject. The remainder of the universal set (30) represents those studying neither.

Problem 2:

Given P(A)=0.6P(A) = 0.6, P(B)=0.5P(B) = 0.5, and P(AβˆͺB)=0.8P(A \cup B) = 0.8, find P(A∩B)P(A \cap B).

Solution:

0.8=0.6+0.5βˆ’P(A∩B)β‡’0.8=1.1βˆ’P(A∩B)β‡’P(A∩B)=0.30.8 = 0.6 + 0.5 - P(A \cap B) \Rightarrow 0.8 = 1.1 - P(A \cap B) \Rightarrow P(A \cap B) = 0.3.

Explanation:

Use the General Addition Rule: P(AβˆͺB)=P(A)+P(B)βˆ’P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). Substitute the known values and solve for the unknown intersection.

Problem 3:

In a Venn diagram, n(A)=10n(A) = 10, n(B)=12n(B) = 12, and n(A∩B)=4n(A \cap B) = 4. If a student is selected from set B, what is the probability they are also in set A?

Solution:

P(A∣B)=n(A∩B)n(B)=412=13P(A | B) = \frac{n(A \cap B)}{n(B)} = \frac{4}{12} = \frac{1}{3}.

Explanation:

This is a conditional probability problem. Since we are told the student is 'from set B', the denominator becomes the total count of B, not the universal set. The numerator is the intersection of A and B.