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Probability - Conditional Probability

Grade 10IGCSE

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 occurring, given that another event has already occurred.

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Notation: P(A∣B)P(A|B) represents the probability of event A happening given that B has happened.

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Restricted Sample Space: In conditional probability, the 'total' denominator changes from the universal set to the size of the given condition (the set B).

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Dependent Events: Two events are dependent if the occurrence of one affects the probability of the other. Most conditional probability problems involve dependent events.

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Tree Diagrams: Used to represent conditional probabilities in multi-stage experiments, especially 'without replacement' scenarios.

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Independence Test: If P(A∣B)=P(A)P(A|B) = P(A), then events A and B are independent.

📐Formulae

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

P(B∣A)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}

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

For Independent Events: P(A∣B)=P(A)P(A|B) = P(A)

💡Examples

Problem 1:

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

Solution:

P(Blue2∣Red1)=37P(Blue_{2} | Red_{1}) = \frac{3}{7}

Explanation:

Initially, there are 8 balls. If the first ball drawn is red, there are now only 7 balls left in the bag. The number of blue balls remains 3. Therefore, the probability of picking a blue ball out of the remaining 7 is 3/7.

Problem 2:

In a group of 100 students, 60 study Math, 40 study Physics, and 20 study both. A student is chosen at random. Given that the student studies Math, what is the probability that they also study Physics?

Solution:

P(Physics∣Math)=2060=13P(Physics | Math) = \frac{20}{60} = \frac{1}{3}

Explanation:

We are given that the student is in the 'Math' group (60 students). Out of these 60, only 20 study Physics (the intersection). Using the formula P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}, we get 20/60.

Problem 3:

For two events AA and BB, P(A)=0.5P(A) = 0.5, P(B)=0.3P(B) = 0.3 and P(A∪B)=0.6P(A \cup B) = 0.6. Find P(A∣B)P(A|B).

Solution:

P(A∣B)=0.20.3=23P(A|B) = \frac{0.2}{0.3} = \frac{2}{3}

Explanation:

First, find P(A∩B)P(A \cap B) using the addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). So, 0.6=0.5+0.3−P(A∩B)⇒P(A∩B)=0.20.6 = 0.5 + 0.3 - P(A \cap B) \Rightarrow P(A \cap B) = 0.2. Then apply the conditional formula: P(A∣B)=P(A∩B)/P(B)=0.2/0.3P(A|B) = P(A \cap B) / P(B) = 0.2 / 0.3.