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Statistics and Probability - Concepts of Probability

Grade 11IB

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

🔑Concepts

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Sample Space and Events: The sample space SS is the set of all possible outcomes of an experiment. An event AA is a subset of the sample space. Visually, think of SS as a rectangular box (the universal set) and AA as a closed shape or circle inside that box containing specific outcomes.

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Complementary Events: The complement of event AA, denoted as A′A' or AcA^c, consists of all outcomes in the sample space SS that are not in AA. Visually, if AA is a circle inside a rectangle, A′A' represents the entire area outside that circle but still within the rectangle. The sum of probabilities is always P(A)+P(A′)=1P(A) + P(A') = 1.

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Venn Diagrams and Set Notation: Venn diagrams use overlapping circles to represent relationships between events. The intersection A∩BA \cap B is the overlapping central region where both events occur. The union A∪BA \cup B covers the total area of both circles combined. The region where only AA occurs is A∩B′A \cap B', represented by the crescent shape of circle AA excluding the overlap.

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Mutually Exclusive Events: Two events are mutually exclusive if they cannot happen at the same time. In a Venn diagram, this is represented by two circles that do not touch or overlap at any point. For these events, the intersection is empty, meaning P(A∩B)=0P(A \cap B) = 0.

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Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other. Unlike mutually exclusive events, independent events usually overlap in a Venn diagram. Mathematically, they satisfy the product rule P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

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Conditional Probability: This measures the probability of an event AA occurring given that event BB has already occurred, denoted P(A∣B)P(A|B). Visually, this 'shrinks' the sample space from the entire rectangle SS down to just the circle BB. We are then looking for the proportion of circle BB that is occupied by the intersection A∩BA \cap B.

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Tree Diagrams: Tree diagrams are used to visualize multi-stage experiments. Each branch represents a possible outcome, with the probability written along the line. To find the probability of a specific path (sequence of events), multiply the probabilities along the branches. To find the total probability of an outcome that occurs at the end of multiple paths, add those path probabilities together.

📐Formulae

P(A)=n(A)n(S)P(A) = \frac{n(A)}{n(S)}

P(A′)=1−P(A)P(A') = 1 - P(A)

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

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

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

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (if AA and BB are independent)

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) (if AA and BB are mutually exclusive)

💡Examples

Problem 1:

In a class of 30 students, 18 study Physics (PP), 15 study Chemistry (CC), and 8 study both. Find the probability that a randomly selected student studies Physics or Chemistry, but not both.

Solution:

  1. Identify the given values: n(S)=30n(S) = 30, n(P)=18n(P) = 18, n(C)=15n(C) = 15, n(P∩C)=8n(P \cap C) = 8.
  2. Find the number of students who study only Physics: n(P)−n(P∩C)=18−8=10n(P) - n(P \cap C) = 18 - 8 = 10.
  3. Find the number of students who study only Chemistry: n(C)−n(P∩C)=15−8=7n(C) - n(P \cap C) = 15 - 8 = 7.
  4. Add the 'only' groups to find 'Physics or Chemistry but not both': 10+7=1710 + 7 = 17.
  5. Calculate the probability: P(Exactly One)=1730P(\text{Exactly One}) = \frac{17}{30}.

Explanation:

This problem uses the principles of a Venn diagram. We subtract the intersection from each individual set to isolate students taking only one subject, ensuring we don't double-count the students in the overlap.

Problem 2:

Two events AA and BB are such that P(A)=0.6P(A) = 0.6, P(B)=0.3P(B) = 0.3, and P(A∪B)=0.72P(A \cup B) = 0.72. Determine if events AA and BB are independent.

Solution:

  1. Use the addition rule to find P(A∩B)P(A \cap B): P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) 0.72=0.6+0.3−P(A∩B)0.72 = 0.6 + 0.3 - P(A \cap B) 0.72=0.9−P(A∩B)0.72 = 0.9 - P(A \cap B) P(A∩B)=0.9−0.72=0.18P(A \cap B) = 0.9 - 0.72 = 0.18
  2. Calculate the product of individual probabilities for the independence test: P(A)×P(B)=0.6×0.3=0.18P(A) \times P(B) = 0.6 \times 0.3 = 0.18
  3. Compare P(A∩B)P(A \cap B) with P(A)×P(B)P(A) \times P(B): Since 0.18=0.180.18 = 0.18, the condition P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) is satisfied.

Explanation:

To check for independence, we first calculate the intersection using the general addition rule. We then check if the intersection equals the product of the individual probabilities. If they are equal, the events are independent.