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Statistics and Probability - Dependent and independent events-extended

Grade 9IB

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

🔑Concepts

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Independent Events: Two events AA and BB are independent if the occurrence of one does not affect the probability of the other. For example, tossing a coin twice.

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Dependent Events: Two events are dependent if the outcome of the first event affects the probability of the second event. This typically occurs in 'without replacement' scenarios.

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Conditional Probability: The probability of event BB occurring given that event AA has already occurred is written as P(B∣A)P(B|A). For independent events, P(B∣A)=P(B)P(B|A) = P(B).

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Product Rule: To find the probability of both events AA and BB occurring (P(A∩B)P(A \cap B)), we multiply their probabilities.

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Tree Diagrams: A visual tool to map out all possible outcomes of a multi-stage experiment. The probability of a specific path is found by multiplying the probabilities along the branches.

📐Formulae

P(A∩B)=P(A)×P(B)(for independent events)P(A \cap B) = P(A) \times P(B) \quad \text{(for independent events)}

P(A∩B)=P(A)×P(B∣A)(for dependent events)P(A \cap B) = P(A) \times P(B|A) \quad \text{(for dependent events)}

P(B∣A)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}

P(A′)=1−P(A)P(A') = 1 - P(A)

💡Examples

Problem 1:

A bag contains 5 red balls and 3 blue balls. If two balls are drawn at random without replacement, find the probability that both balls are red.

Solution:

Let R1R_1 be the event that the first ball is red, and R2R_2 be the event that the second ball is red.

P(R1)=58P(R_1) = \frac{5}{8}

Since the first ball is not replaced, there are now 4 red balls left out of a total of 7 balls.

P(R2∣R1)=47P(R_2|R_1) = \frac{4}{7}

Using the product rule for dependent events:

P(R1∩R2)=P(R1)×P(R2∣R1)P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1)

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}

The probability is 514\frac{5}{14}.

Explanation:

Because the first ball is not replaced, the total number of balls and the number of red balls decrease, making the events dependent.

Problem 2:

A fair six-sided die is rolled and a coin is flipped. What is the probability of rolling a number greater than 4 and flipping a 'Tails'?

Solution:

Let AA be the event of rolling a number greater than 4 (55 or 66), and BB be the event of flipping Tails.

P(A)=26=13P(A) = \frac{2}{6} = \frac{1}{3}

P(B)=12P(B) = \frac{1}{2}

Since the die roll and the coin flip do not affect each other, the events are independent.

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

P(A∩B)=13×12=16P(A \cap B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}

The probability is 16\frac{1}{6}.

Explanation:

The outcomes of a die and a coin are independent, so we simply multiply their individual probabilities.

Problem 3:

In a class of 30 students, 18 study Music (MM) and 12 study Art (AA). 5 students study both. If a student is chosen at random and found to study Music, what is the probability they also study Art?

Solution:

We are looking for the conditional probability P(A∣M)P(A|M).

From the data: P(M)=1830P(M) = \frac{18}{30} P(A∩M)=530P(A \cap M) = \frac{5}{30}

Using the formula for conditional probability:

P(A∣M)=P(A∩M)P(M)P(A|M) = \frac{P(A \cap M)}{P(M)}

P(A∣M)=5301830=518P(A|M) = \frac{\frac{5}{30}}{\frac{18}{30}} = \frac{5}{18}

The probability is 518\frac{5}{18}.

Explanation:

This is a conditional probability problem where the sample space is restricted to only those students who study Music.