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Statistics and Probability - Combined, mutually exclusive, and independent events

Grade 10IB

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

🔑Concepts

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Sample Space and Probability: The sample space, denoted by UU or SS, represents the set of all possible outcomes of an experiment. Visually, this is the rectangular boundary of a Venn diagram. The probability of an event AA is the ratio of the number of favorable outcomes to the total outcomes in the sample space: P(A)=n(A)n(U)P(A) = \frac{n(A)}{n(U)}.

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Combined Events (The Union): The union of two events AA and BB (written as A∪BA \cup B) consists of all outcomes that are in AA, or in BB, or in both. In a Venn diagram, this is represented by shading the entire area covered by both circles. To calculate this, we use 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).

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Mutually Exclusive Events: These are events that cannot occur at the same time. Visually, in a Venn diagram, mutually exclusive events are shown as two disjoint circles that do not overlap. Because they have no common outcomes, the probability of their intersection is zero: 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 occurring. For example, rolling a die and then flipping a coin. Visually, independent events are often represented using a Tree Diagram, where the probabilities on the second set of branches remain the same regardless of the outcome of the first branch.

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Intersection of Events: The intersection of events AA and BB (written as A∩BA \cap B) consists of outcomes that are common to both events. On a Venn diagram, this is the 'overlap' or the football-shaped region where the two circles meet. For independent events, the intersection is calculated by multiplying their individual probabilities: P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

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Complementary Events: The complement of event AA (written as A′A') consists of all outcomes in the sample space that are not in AA. Visually, if event AA is a circle, A′A' is everything inside the rectangle but outside that circle. The sum of the probability of an event and its complement is always 11: P(A)+P(A′)=1P(A) + P(A') = 1.

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Tree Diagrams for Combined Events: A tree diagram is a visual tool used to map out outcomes of multiple stages of an experiment. Each 'branch' represents a possible outcome, and the probability is written along the branch. To find the probability of a specific path (a combined event), you multiply the probabilities along the branches. To find the total probability of several paths, you add the resulting products.

📐Formulae

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

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

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

P(A′)=1−P(A)P(A') = 1 - P(A)

P(A∩B)=0P(A \cap B) = 0 (for mutually exclusive events)

💡Examples

Problem 1:

In a group of 30 students, 18 study Spanish, 12 study French, and 5 study both languages. If a student is chosen at random, find the probability that they study Spanish or French.

Solution:

  1. Identify the given values: n(S)=18n(S) = 18, n(F)=12n(F) = 12, n(S∩F)=5n(S \cap F) = 5, and total n(U)=30n(U) = 30. \ 2. Convert these to probabilities: P(S)=1830P(S) = \frac{18}{30}, P(F)=1230P(F) = \frac{12}{30}, and P(S∩F)=530P(S \cap F) = \frac{5}{30}. \ 3. Use the Addition Rule for combined events: P(S∪F)=P(S)+P(F)−P(S∩F)P(S \cup F) = P(S) + P(F) - P(S \cap F). \ 4. Substitute values: P(S∪F)=1830+1230−530=2530P(S \cup F) = \frac{18}{30} + \frac{12}{30} - \frac{5}{30} = \frac{25}{30}. \ 5. Simplify: P(S∪F)=56P(S \cup F) = \frac{5}{6}.

Explanation:

Since some students study both languages (the events are not mutually exclusive), we must subtract the intersection to avoid double-counting those students when calculating the union.

Problem 2:

A bag contains 4 red marbles and 6 blue marbles. A marble is drawn, its color is recorded, it is replaced, and then a second marble is drawn. Find the probability of drawing a red marble and then a blue marble.

Solution:

  1. Determine the probability of drawing a red marble: P(R)=44+6=410=0.4P(R) = \frac{4}{4+6} = \frac{4}{10} = 0.4. \ 2. Since the marble is replaced, the events are independent. The probability of drawing a blue marble remains: P(B)=610=0.6P(B) = \frac{6}{10} = 0.6. \ 3. Use the Multiplication Rule for independent events: P(R∩B)=P(R)×P(B)P(R \cap B) = P(R) \times P(B). \ 4. Calculate: P(R∩B)=0.4×0.6=0.24P(R \cap B) = 0.4 \times 0.6 = 0.24.

Explanation:

Because the first marble is replaced, the second draw is independent of the first. We multiply the probabilities of the two specific outcomes to find the probability of the combined event occurring in that sequence.