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Statistics and Probability - Concepts of probability (sample space, events)

Grade 10IB

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

🔑Concepts

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A Random Experiment is a process where the result is uncertain. The Sample Space, denoted by SS or UU, is the set of all possible outcomes of an experiment. Visually, this can be represented as a rectangular boundary in a Venn diagram or a list of items inside curly brackets, such as S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\} for a standard six-sided die.

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An Event is a specific outcome or a collection of outcomes that is a subset of the sample space. In a Venn diagram, an event is typically shown as a labeled circle inside the sample space rectangle. If the event is 'rolling an even number', it would include the outcomes {2,4,6}\{2, 4, 6\}.

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The Probability Scale measures the likelihood of an event occurring, ranging from 00 to 11. Visually, this is a horizontal line where 00 represents an 'Impossible' event, 11 represents a 'Certain' event, and 0.50.5 represents an 'Equally Likely' chance (like a coin flip). Values closer to 11 indicate higher likelihood.

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Complementary Events represent the probability of an event NOT occurring, denoted as A′A' or AcA^c. In a Venn diagram, if AA is a circle, A′A' is the entire shaded area outside that circle but within the universal rectangle. The sum of the probability of an event and its complement is always 11.

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Mutually Exclusive Events are events that cannot happen at the same time. For example, rolling a 22 and a 55 on a single die are mutually exclusive. Visually, these are represented in a Venn diagram as two separate circles that do not overlap or intersect.

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Tree Diagrams are visual tools used to map out the sample space of multi-stage experiments. They consist of 'branches' stemming from a common point for the first event, with subsequent branches growing from the ends of the first set for the second event. Each branch is labeled with its outcome and probability; the total probability at the end of each path is found by multiplying along the branches.

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Theoretical Probability assumes all outcomes in a sample space are equally likely. It is calculated by dividing the number of successful outcomes by the total number of possible outcomes in the sample space.

📐Formulae

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

P(A)+P(A′)=1P(A) + P(A') = 1

P(A′)=1−P(A)P(A') = 1 - P(A)

0≤P(A)≤10 \le P(A) \le 1

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

For mutually exclusive events: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

💡Examples

Problem 1:

A fair six-sided die is rolled once. Let AA be the event of rolling a prime number. List the sample space SS, the event AA, and calculate P(A)P(A).

Solution:

  1. List the sample space: S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}. The total number of outcomes is n(S)=6n(S) = 6.
  2. Identify the prime numbers in the sample space: A={2,3,5}A = \{2, 3, 5\}. The number of successful outcomes is n(A)=3n(A) = 3.
  3. Apply the probability formula: P(A)=n(A)n(S)=36P(A) = \frac{n(A)}{n(S)} = \frac{3}{6}.
  4. Simplify the fraction: P(A)=12P(A) = \frac{1}{2} or 0.50.5.

Explanation:

We first identify all possible results of the die roll to establish the denominator. Then, we identify which of those results satisfy the condition of being a 'prime number' (note that 11 is not prime) to establish the numerator.

Problem 2:

A bag contains 55 red marbles, 33 blue marbles, and 22 green marbles. If one marble is drawn at random, find the probability that it is NOT blue.

Solution:

  1. Calculate the total number of marbles: n(S)=5+3+2=10n(S) = 5 + 3 + 2 = 10.
  2. Let BB be the event that the marble is blue. The number of blue marbles is n(B)=3n(B) = 3.
  3. Calculate P(B)=n(B)n(S)=310P(B) = \frac{n(B)}{n(S)} = \frac{3}{10}.
  4. Use the complement rule to find the probability of NOT blue (B′B'): P(B′)=1−P(B)P(B') = 1 - P(B).
  5. P(B′)=1−310=710P(B') = 1 - \frac{3}{10} = \frac{7}{10} or 0.70.7.

Explanation:

The problem asks for the probability of the complement of 'blue'. We can either sum the probabilities of red and green or subtract the probability of blue from the total probability of 11.