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Probability - Relative Frequency and Theoretical Probability

Grade 12IGCSE

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 an ideal scenario where all outcomes are equally likely.

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

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

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Sample Space: The set of all possible outcomes of a probability experiment, denoted by S.

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Mutually Exclusive Events: Events that cannot happen at the same time. For these events, P(A or B) = P(A) + P(B).

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Independent Events: The occurrence of one event does not affect the probability of the other. For these events, P(A and B) = P(A) × P(B).

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Expected Frequency: The number of times an event is predicted to occur over a specific number of trials, calculated as n × P(A).

📐Formulae

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

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

P(A′)=1−P(A) (Complementary Events)P(A') = 1 - P(A) \text{ (Complementary Events)}

P(A∪B)=P(A)+P(B)−P(A∩B) (Addition Rule)P(A \cup B) = P(A) + P(B) - P(A \cap B) \text{ (Addition Rule)}

Expected value=n×P(A)\text{Expected value} = n \times P(A)

P(A∩B)=P(A)×P(B∣A) (General Multiplication Rule)P(A \cap B) = P(A) \times P(B|A) \text{ (General Multiplication Rule)}

💡Examples

Problem 1:

A bag contains 5 red marbles and 3 blue marbles. A marble is drawn, its color recorded, and it is NOT replaced. A second marble is then drawn. Find the theoretical probability that both marbles are red.

Solution:

P(R1∩R2)=58×47=2056=514P(R_1 \cap R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

Explanation:

This is a conditional probability problem. The first draw has 5 red marbles out of 8. Since we do not replace it, the second draw has only 4 red marbles left out of a total of 7 marbles.

Problem 2:

A biased six-sided die is rolled 500 times. The number '6' appears 125 times. (a) Calculate the relative frequency of rolling a 6. (b) If the die is rolled 2000 times, how many 6s would you expect to see based on this experiment?

Solution:

(a) RF=125500=0.25\text{RF} = \frac{125}{500} = 0.25. (b) Expected=2000×0.25=500\text{Expected} = 2000 \times 0.25 = 500.

Explanation:

Relative frequency is calculated from observed data (125/500). To find the expected frequency for a larger sample, multiply the experimental probability (0.25) by the new number of trials (2000).

Problem 3:

In a group of 100 students, 60 study Math, 40 study Physics, and 20 study both. Find the probability that a randomly selected student studies neither Math nor Physics.

Solution:

P(M∪P)=P(M)+P(P)−P(M∩P)=0.60+0.40−0.20=0.80P(M \cup P) = P(M) + P(P) - P(M \cap P) = 0.60 + 0.40 - 0.20 = 0.80. Therefore, P(neither)=1−0.80=0.20P(\text{neither}) = 1 - 0.80 = 0.20.

Explanation:

First, use the Addition Rule to find the probability that a student studies at least one of the subjects. Subtract the intersection (those who study both) to avoid double-counting. The probability of 'neither' is the complement of 'either'.