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Probability - Combined events

Grade 11IGCSE

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

🔑Concepts

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Sample Space: The set of all possible outcomes of an experiment, often represented using a list, table, or Venn diagram.

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Independent Events: Events where the outcome of the first does not affect the outcome of the second.

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Dependent Events: Events where the outcome of the first event changes the probability of the second event (often 'without replacement' scenarios).

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Mutually Exclusive Events: Events that cannot happen at the same time; their intersection is zero.

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Tree Diagrams: A visual tool to represent sequences of events where probabilities are multiplied along branches and added across different paths.

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Possibility Space Diagrams: A grid/table used to show all possible outcomes for two combined events, such as rolling two dice.

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The 'At Least' Rule: Using the complement to solve complex problems, where P(extatleastone)=1−P(extnone)P( ext{at least one}) = 1 - P( ext{none}).

📐Formulae

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

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) (For Mutually Exclusive Events)

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (For Independent Events)

P(A′)=1−P(A)P(A') = 1 - P(A) (Complementary Events)

P(B∣A)=P(A∩B)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)} (Conditional Probability - Higher Tier focus)

💡Examples

Problem 1:

A bag contains 5 red balls and 3 blue balls. Two balls are taken at random from the bag without replacement. Find the probability that both balls are red.

Solution:

58×47=2056=514\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

Explanation:

This is a dependent event. For the first draw, the probability of red is 5/8. Since the ball is not replaced, there are now 4 red balls left out of a total of 7. We multiply the probabilities along the branch.

Problem 2:

A fair six-sided die is rolled and a fair coin is tossed. What is the probability of rolling a number greater than 4 and getting a Head?

Solution:

P(>4)=26=13P(>4) = \frac{2}{6} = \frac{1}{3}; P(H)=12P(H) = \frac{1}{2}. P(>4 and H)=13×12=16P(>4 \text{ and } H) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}

Explanation:

These are independent events because the result of the die does not affect the coin. We find the individual probabilities and multiply them.

Problem 3:

In a class of 30 students, 18 study Biology (B), 15 study Chemistry (C), and 7 study both. A student is picked at random. Find the probability they study Biology or Chemistry but not both.

Solution:

P(B only)=18−730=1130P(B \text{ only}) = \frac{18-7}{30} = \frac{11}{30}; P(C only)=15−730=830P(C \text{ only}) = \frac{15-7}{30} = \frac{8}{30}. Total = 11+830=1930\frac{11+8}{30} = \frac{19}{30}

Explanation:

Using a Venn diagram approach, we subtract the intersection (7) from the individual totals to find those who study only one subject, then add those mutually exclusive probabilities.