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Probability - Theoretical and relative frequency

Grade 11IGCSE

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

🔑Concepts

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Theoretical Probability: The likelihood of an event occurring based on all possible outcomes in a perfectly fair scenario.

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Relative Frequency (Experimental Probability): The ratio of the number of times an event occurs to the total number of trials conducted.

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Sample Space: The set of all possible outcomes of an experiment, usually denoted by S.

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Mutually Exclusive Events: Events that cannot happen at the same time; P(A∩B)=0P(A \cap B) = 0.

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Complementary Events: The probability that an event does not occur, calculated as P(A′)=1−P(A)P(A') = 1 - P(A).

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Law of Large Numbers: As the number of trials increases, the relative frequency tends to get closer to the theoretical probability.

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Expected Frequency: The number of times an event is predicted to occur over a specific number of trials.

📐Formulae

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

Relative Frequency=Frequency of the eventTotal number of trials\text{Relative Frequency} = \frac{\text{Frequency of the event}}{\text{Total number of trials}}

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

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

Expected Frequency=n×P(A)\text{Expected Frequency} = n \times P(A), where nn is the number of trials.

💡Examples

Problem 1:

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

Solution:

P(Prime)=36=0.5P(\text{Prime}) = \frac{3}{6} = 0.5

Explanation:

The sample space is {1, 2, 3, 4, 5, 6}. The prime numbers in this set are {2, 3, 5}. There are 3 favorable outcomes out of 6 possible outcomes.

Problem 2:

A spinner is spun 200 times. The color 'Red' appears 46 times. Calculate the relative frequency of landing on Red.

Solution:

RelativeFrequency=46200=0.23Relative Frequency = \frac{46}{200} = 0.23

Explanation:

Relative frequency is calculated by dividing the observed frequency of the event (46) by the total number of trials (200).

Problem 3:

The probability that a seed germinates is 0.85. If a farmer plants 1,200 seeds, how many are expected to germinate?

Solution:

ExpectedFrequency=1200×0.85=1,020Expected Frequency = 1200 \times 0.85 = 1,020

Explanation:

To find the expected frequency, multiply the total number of trials (n = 1200) by the theoretical probability (p = 0.85).