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Probability - Conditional Probability

Grade 12IGCSE

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

🔑Concepts

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Definition: Conditional probability is the probability of an event A occurring given that event B has already occurred.

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Restricted Sample Space: Conditional probability effectively reduces the sample space to only the outcomes where event B is true.

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Dependent vs. Independent Events: If P(A|B) = P(A), then events A and B are independent.

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The Multiplication Rule: Derived from the conditional formula, used to find the probability of the intersection of two events.

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Tree Diagrams: A visual method where the second set of branches represents conditional probabilities.

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Bayes' Theorem (Introductory): Calculating P(B|A) when P(A|B) is known.

📐Formulae

P(A∣B)=P(A∩B)P(B), where P(B)>0P(A|B) = \frac{P(A \cap B)}{P(B)}, \text{ where } P(B) > 0

P(A∩B)=P(B)×P(A∣B)P(A \cap B) = P(B) \times P(A|B)

P(A∣B)=P(A) (Condition for Independence)P(A|B) = P(A) \text{ (Condition for Independence)}

P(A′∣B)=1−P(A∣B)P(A'|B) = 1 - P(A|B)

💡Examples

Problem 1:

In a class of 30 students, 18 study Physics, 15 study Biology, and 8 study both. If a student is chosen at random and it is known they study Biology, what is the probability they also study Physics?

Solution:

P(P∣B)=n(P∩B)n(B)=815P(P|B) = \frac{n(P \cap B)}{n(B)} = \frac{8}{15}

Explanation:

We are given that the student is in the 'Biology' group (15 students). Within that group, 8 students also study Physics. Therefore, the probability is the intersection divided by the condition.

Problem 2:

A bag contains 5 red and 3 blue marbles. Two marbles are drawn one after another without replacement. Find the probability that the second marble is blue, given that the first marble was red.

Solution:

P(B2∣R1)=37P(B_2|R_1) = \frac{3}{7}

Explanation:

Initially, there are 8 marbles. After one red marble is removed, 7 marbles remain in the bag. The number of blue marbles is still 3. Thus, the probability of picking a blue marble from the remaining set is 3/7.

Problem 3:

Given P(A)=0.6P(A) = 0.6, P(B)=0.5P(B) = 0.5, and P(A∪B)=0.8P(A \cup B) = 0.8. Find P(B∣A)P(B|A).

Solution:

P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.5−0.8=0.3P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3. Then, P(B∣A)=P(A∩B)P(A)=0.30.6=0.5P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.6} = 0.5.

Explanation:

First, use the Addition Rule to find the probability of the intersection. Then, apply the conditional probability formula by dividing the intersection by the probability of the given condition (A).