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Probability - Conditional probability

Grade 11IGCSE

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) denotes the probability of event A occurring given that event B has happened.

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Restricted Sample Space: In conditional probability, the 'universe' of outcomes is reduced to only those outcomes within event B.

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Without Replacement: Common in IGCSE problems involving bags of items; the outcome of the first draw changes the probabilities for the second draw.

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Independence: Two events are independent if P(A∣B)=P(A)P(A|B) = P(A), meaning the occurrence of B does not affect the probability of A.

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Tree Diagrams: Used to visualize sequential events where the second set of branches represents conditional probabilities.

📐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)}

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

For independent events: P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

💡Examples

Problem 1:

A bag contains 5 red marbles 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 drawn was red.

Solution:

37\frac{3}{7}

Explanation:

Initially, there are 8 marbles. If the first marble drawn is red, there are now 7 marbles remaining in the bag. Since the first marble was red, all 3 blue marbles are still in the bag. Therefore, the probability of picking a blue marble given the first was red is 3 out of 7.

Problem 2:

In a group of 100 students, 60 study Biology, 40 study Chemistry, and 20 study both. If a student is chosen at random and it is known they study Biology, what is the probability they also study Chemistry?

Solution:

13\frac{1}{3}

Explanation:

Let B be the event 'studies Biology' and C be 'studies Chemistry'. We are looking for P(C∣B)P(C|B). Using the formula P(C∣B)=P(C∩B)P(B)P(C|B) = \frac{P(C \cap B)}{P(B)}. Here, n(B)=60n(B) = 60 and n(C∩B)=20n(C \cap B) = 20. So, P(C∣B)=2060=13P(C|B) = \frac{20}{60} = \frac{1}{3}.

Problem 3:

Given that 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, calculate P(A∣B)P(A|B).

Solution:

23\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). This gives 0.6=0.5+0.3−P(A∩B)0.6 = 0.5 + 0.3 - P(A \cap B), so P(A∩B)=0.2P(A \cap B) = 0.2. Now, apply the conditional formula: P(A∣B)=P(A∩B)P(B)=0.20.3=23P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = \frac{2}{3}.