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Statistics and Probability - Sample Space Diagrams for Single Events

Grade 7IB

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

🔑Concepts

The Sample Space (SS) is the set of all possible outcomes of a probability experiment. For a single event, it is often visually represented as a list within curly brackets, such as S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\} for a standard die, or as a set of points on a number line.

The Probability Scale is a visual representation of the likelihood of an event, ranging from 00 (Impossible) to 11 (Certain). A probability of 0.50.5 (or 12\frac{1}{2}) represents an 'Even Chance,' such as flipping a tail on a fair coin.

Equally Likely Outcomes occur when every possible result in the sample space has the same chance of occurring. Visually, this is represented by a fair spinner divided into equal-sized sectors or a balanced die where each face has the same surface area.

An Event is a specific outcome or a collection of outcomes from the sample space. For example, in the sample space of a die, 'rolling an even number' is an event consisting of the subset {2,4,6}\{2, 4, 6\}.

The Complement of an Event (AA') includes all outcomes in the sample space that are not part of event AA. If you visualize the sample space as a box (Venn Diagram), if event AA is a circle inside, the complement AA' is the entire area outside that circle but within the box.

Theoretical Probability is determined by mathematical reasoning rather than physical experiments. It assumes that physical objects like coins and dice are 'fair' and 'unbiased,' meaning their visual and physical symmetry ensures equal probability for all faces.

The Sum of Probabilities for all possible mutually exclusive outcomes in a single-event sample space must always equal 11. This can be visualized as a pie chart where the sum of all individual slices accounts for the whole circle.

📐Formulae

P(E)=Number of favorable outcomesTotal number of possible outcomesP(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

P(A)+P(A)=1P(A) + P(A') = 1

P(A)=1P(A)P(A') = 1 - P(A)

0P(E)10 \le P(E) \le 1

💡Examples

Problem 1:

A fair spinner is divided into 88 equal sections numbered 11 to 88. What is the probability of spinning a prime number?

Solution:

  1. List the sample space: S={1,2,3,4,5,6,7,8}S = \{1, 2, 3, 4, 5, 6, 7, 8\}. The total number of outcomes n(S)=8n(S) = 8.
  2. Identify the prime numbers in the sample space: {2,3,5,7}\{2, 3, 5, 7\}. Note that 11 is not a prime number.
  3. Count the favorable outcomes: n(E)=4n(E) = 4.
  4. Use the formula: P(prime)=n(E)n(S)=48P(\text{prime}) = \frac{n(E)}{n(S)} = \frac{4}{8}.
  5. Simplify the fraction: 12\frac{1}{2} or 0.50.5.

Explanation:

First, we identify all possible results on the spinner. Then, we isolate the results that fit the criteria (prime numbers) and divide that count by the total number of sections.

Problem 2:

A bag contains 44 red, 33 green, and 55 blue marbles. If one marble is selected at random, what is the probability that it is NOT red?

Solution:

  1. Calculate the total number of outcomes: 4+3+5=124 + 3 + 5 = 12.
  2. Find the number of red marbles: n(red)=4n(\text{red}) = 4.
  3. Calculate the probability of selecting a red marble: P(red)=412=13P(\text{red}) = \frac{4}{12} = \frac{1}{3}.
  4. Use the complement rule to find the probability of NOT red: P(red)=1P(red)P(\text{red}') = 1 - P(\text{red}).
  5. P(red)=113=23P(\text{red}') = 1 - \frac{1}{3} = \frac{2}{3}.

Explanation:

To find the probability of an event not happening, we can subtract the probability of it happening from 11. Alternatively, we could have added the green and blue marbles together (3+5=83 + 5 = 8) and divided by 1212 to get 812=23\frac{8}{12} = \frac{2}{3}.