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Statistics and Probability - Introduction to Probability

Grade 8IB

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

🔑Concepts

The Probability Scale: Probability is a measure of the likelihood that an event will occur, represented on a scale from 00 to 11. Visually, this is a horizontal line where 00 (impossible) is on the far left, 0.50.5 (even chance) is in the middle, and 11 (certain) is on the far right. Values closer to 11 represent 'likely' events, while values closer to 00 represent 'unlikely' events.

Theoretical Probability: This is calculated based on reasoning and the assumption that all outcomes are equally likely. For example, when looking at a standard six-sided die, we assume each face has a 11 in 66 chance. The sample space is the set of all possible outcomes, often listed in curly brackets as {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.

Experimental Probability (Relative Frequency): This is based on actual data or experiments. It is calculated after performing trials. For instance, if you flip a coin 100100 times and get heads 5555 times, the experimental probability is 0.550.55. As the number of trials increases, the experimental probability usually gets closer to the theoretical probability.

Sample Space Diagrams: For experiments involving two steps (like rolling two dice), a sample space diagram or a 2D grid is used. Imagine a square grid where the horizontal axis represents the outcomes of the first die (161-6) and the vertical axis represents the second die (161-6). Each intersection point in the grid represents a unique combined outcome, such as (2,3)(2, 3), totaling 3636 possible outcomes.

Complementary Events: The complement of an event AA is the event that AA does not happen, denoted as AA'. Visually, if you imagine a Venn diagram where a circle represents event AA inside a rectangular box (the universal set), the area outside the circle but inside the box represents AA'. The sum of the probability of an event and its complement is always 11.

Tree Diagrams: These are used to visualize probabilities of multi-stage events. Each stage of the experiment is represented by a set of 'branches' growing out from a single point. For each branch, the outcome is written at the end and the probability is written along the line. To find the probability of a specific path, you multiply the probabilities along those branches.

Equally Likely Outcomes: In many theoretical models, we assume outcomes are 'fair'. This means no outcome is weighted more heavily than another. If a spinner is divided into four equal-sized colored quadrants (red, blue, yellow, green), the visual symmetry ensures each color has a probability of 14\frac{1}{4}.

📐Formulae

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

Relative Frequency=Frequency of an outcomeTotal number of trials\text{Relative Frequency} = \frac{\text{Frequency of an outcome}}{\text{Total number of trials}}

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

P(A or B)=P(A)+P(B) (for mutually exclusive events)P(A \text{ or } B) = P(A) + P(B) \text{ (for mutually exclusive events)}

P(all outcomes)=1\sum P(\text{all outcomes}) = 1

💡Examples

Problem 1:

A fair six-sided die is rolled. What is the probability of rolling a prime number?

Solution:

  1. Identify the sample space SS: S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}. The total number of outcomes is 66.
  2. Identify the favorable outcomes (prime numbers): Prime numbers on a die are 2,3,2, 3, and 55. Note that 11 is not a prime number.
  3. The number of favorable outcomes is 33.
  4. Apply the formula: P(prime)=36P(\text{prime}) = \frac{3}{6}.
  5. Simplify: P(prime)=12P(\text{prime}) = \frac{1}{2} or 0.50.5.

Explanation:

To solve this, we list all possible results of the die roll and count how many satisfy the specific condition (being a prime number), then divide by the total.

Problem 2:

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

Solution:

  1. Calculate the total number of marbles: 5+3+2=105 + 3 + 2 = 10.
  2. Identify the probability of picking a blue marble: P(blue)=310P(\text{blue}) = \frac{3}{10}.
  3. Use the complement rule to find the probability of 'not blue': P(not blue)=1P(blue)P(\text{not blue}) = 1 - P(\text{blue}).
  4. Substitute the values: P(not blue)=1310P(\text{not blue}) = 1 - \frac{3}{10}.
  5. Calculate the final result: P(not blue)=710P(\text{not blue}) = \frac{7}{10} or 0.70.7.

Explanation:

This problem uses the complement rule P(A)=1P(A)P(A') = 1 - P(A). Alternatively, you could add the probabilities of all non-blue outcomes (red and green), giving 5+210=710\frac{5+2}{10} = \frac{7}{10}.