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Statistics and Probability - Conditional Probability and Independence

Grade 11IB

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

🔑Concepts

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Conditional Probability Definition: The probability of an event AA occurring given that event BB has already occurred is denoted as P(A∣B)P(A|B). Visually, this represents a 'restriction of the sample space' where we ignore everything outside of circle BB in a Venn diagram and focus only on the intersection A∩BA \cap B within that circle.

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Independent Events: Two events are independent if the occurrence of one does not change the probability of the other. In a Venn diagram, there is no visual 'link' between the events, but mathematically it means P(A∣B)=P(A)P(A|B) = P(A). If P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B), the events are independent.

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The Multiplication Rule: For any two events, the probability of both occurring is P(A∩B)=P(B)×P(A∣B)P(A \cap B) = P(B) \times P(A|B). This rule is the basis for Tree Diagrams, where you multiply probabilities along the branches to find the probability of a specific path of outcomes.

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Tree Diagrams: A visual representation of sequences of events. Each set of branches stems from a single node and must sum to 11. To find the probability of a combined outcome, you multiply the probabilities along the path. For example, in a two-stage experiment, the second set of branches represents conditional probabilities based on the first outcome.

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Mutually Exclusive Events: These are events that cannot happen at the same time, meaning P(A∩B)=0P(A \cap B) = 0. In a Venn diagram, these are represented by two disjoint circles that do not overlap. It is important to distinguish this from independence; if two events are mutually exclusive and have non-zero probabilities, they are actually dependent because the occurrence of one makes the probability of the other zero.

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The Complement Rule in Conditional Probability: The probability of 'not AA' given BB is calculated as P(A′∣B)=1−P(A∣B)P(A'|B) = 1 - P(A|B). This is useful in calculations where finding the inverse probability is simpler than finding the direct one.

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Combined Events and Venn Diagrams: To visualize P(A∪B)P(A \cup B), you look at the total area covered by both circles AA and BB. 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) ensures that the overlapping intersection is not counted twice.

📐Formulae

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

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

Independence Test 1: P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

Independence Test 2: P(A∣B)=P(A)P(A|B) = P(A)

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

Complementary Events: P(A′)=1−P(A)P(A') = 1 - P(A)

Total Probability: P(A)=P(A∩B)+P(A∩B′)P(A) = P(A \cap B) + P(A \cap B')

💡Examples

Problem 1:

In a group of 100 students, 70 study Mathematics, 60 study Physics, and 40 study both. A student is selected at random. Find the probability that the student studies Physics, given that they study Mathematics.

Solution:

  1. Identify the given probabilities: P(M)=70100=0.7P(M) = \frac{70}{100} = 0.7, P(Ph)=60100=0.6P(Ph) = \frac{60}{100} = 0.6, and P(M∩Ph)=40100=0.4P(M \cap Ph) = \frac{40}{100} = 0.4. \ 2. Use the conditional probability formula: P(Ph∣M)=P(Ph∩M)P(M)P(Ph|M) = \frac{P(Ph \cap M)}{P(M)}. \ 3. Substitute the values: P(Ph∣M)=0.40.7=47≈0.571P(Ph|M) = \frac{0.4}{0.7} = \frac{4}{7} \approx 0.571.

Explanation:

We are restricting the sample space to only the 70 students who study Mathematics. Among these 70 students, 40 also study Physics. Therefore, the probability is the ratio of those who do both to the total in the 'given' group.

Problem 2:

A bag contains 5 red and 3 blue marbles. Two marbles are drawn one after the other without replacement. Find the probability that both marbles are blue.

Solution:

  1. Let B1B_1 be the event the first marble is blue and B2B_2 be the event the second marble is blue. \ 2. Find P(B1)P(B_1): There are 3 blue out of 8 total marbles, so P(B1)=38P(B_1) = \frac{3}{8}. \ 3. Find P(B2∣B1)P(B_2|B_1): If the first was blue, 7 marbles remain and only 2 are blue. So, P(B2∣B1)=27P(B_2|B_1) = \frac{2}{7}. \ 4. Use the multiplication rule: P(B1∩B2)=P(B1)×P(B2∣B1)P(B_1 \cap B_2) = P(B_1) \times P(B_2|B_1). \ 5. Calculate: 38×27=656=328≈0.107\frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28} \approx 0.107.

Explanation:

This is a dependent event scenario because the first draw changes the composition of the bag for the second draw. We multiply the probability of the first event by the conditional probability of the second event occurring after the first.