Quick Notes

Quick Notes

Mathematics Chapter 9: Probability of Combined Events
9.2 Dependent Events and Independent Events 
9.2 Dependent Events and Independent Events
MATHEMATICS • Form 4 • Chapter 9: Probability of Combined Events

9.2 Dependent Events and Independent Events

Combined events occur when two or more single events are combined together. These combined events can be classified into two main categories: independent events and dependent events. Understanding whether the outcome of one event affects another is crucial in determining the total sample space and selecting the appropriate method—such as the multiplication rule or tree diagrams—to calculate their probabilities.

Learning Objectives

  • Differentiate between dependent events and independent events.
  • Make and verify conjectures about the formula of probability of combined events (multiplication rule: \(P(A \text{ and } B) = P(A) \times P(B)\)).
  • Determine the probability of combined events for both dependent events and independent events by listing outcomes or sketching tree diagrams.

Tree Diagram for Sampling Without Replacement

See how probabilities change at the second branch when items are chosen without replacement, compared with independent events.

Start
\(G\)
\(\frac{8}{9}\)
\(R\)
\(\frac{1}{9}\)
\(G\)
\(\frac{7}{8}\)
\((G, G)\)
\(R\)
\(\frac{1}{8}\)
\((G, R)\)
\(G\)
\(\frac{8}{8}\)
\((R, G)\)
Without replacement: after the first draw, the total number of marbles and the number of green marbles both reduce by 1. The second-branch probabilities change.

Brief Explanation

Independent Events

Event \(A\) and event \(B\) are independent events if the occurrence of event \(A\) has no effect on the occurrence of event \(B\), and vice versa.

Example: choosing a card from a box, recording it, and returning it before choosing the second card. The probability of selecting a particular card in the second draw remains unchanged.

Dependent Events

Event \(A\) and event \(B\) are dependent events if the occurrence of event \(A\) affects the occurrence of event \(B\).

Example: choosing a card or pen from a container one by one without replacement. The reduction in the total number of items and specific items affects the probability of the second selection.

Multiplication Rule of Probability

The probability of the intersection of two independent events \(A\) and \(B\) equals the product of their individual probabilities:

\[P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)\]

For three independent events \(A\), \(B\), and \(C\):

\[P(A \cap B \cap C) = P(A) \times P(B) \times P(C)\]
1
Listing All Possible Outcomes
List all elements in sample space \(S\) using tables or systematic pairs, then count \(n(S)\). Identify outcomes favourable to event \(E\) and count \(n(E)\). Compute \(P(E) = \frac{n(E)}{n(S)}\).
2
Using the Multiplication Rule
Apply \(P(A \text{ and } B) = P(A) \times P(B)\) directly for independent events.
3
Using Tree Diagrams
A tree diagram displays all possible outcomes of an event. Each branch represents an outcome accompanied by its respective probability. For dependent events without replacement, adjust the denominators and numerators along sequential branches to reflect items remaining. Multiply branch probabilities along a specific path to determine the joint probability of that outcome sequence.

Try Answering First

Read the question, answer it in your mind, then press “Check Answer”.

1

Identify whether obtaining two red pens when two pens are drawn one by one without replacement from a box containing 3 red pens and 2 blue pens is a dependent or independent event. Justify your answer.

Answer: Dependent events, because the first red pen chosen is not replaced, which reduces the total number of pens and the number of red pens, thereby affecting the probability of selecting the second red pen.
2

Box \(A\) contains cards numbered \(3, 5, 7, 9\) and Box \(B\) contains cards labelled \(X, Y, Z\). One card is drawn at random from each box. Calculate the probability of getting a factor of 9 from Box \(A\) and the letter ‘\(Z\)’ from Box \(B\).

Answer: Factors of 9 in Box \(A\) are \(\{3, 9\}\), so \(P(\text{factor of } 9) = \frac{2}{4} = \frac{1}{2}\). The probability of getting ‘\(Z\)’ is \(P(Z) = \frac{1}{3}\). Since the events are independent, \(P(\text{factor of } 9 \text{ and } Z) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\).
3

A bag contains 8 green marbles and 1 red marble. Two marbles are selected one by one without replacement. Find the probability that the second marble chosen is red.

Answer: The second marble can be red only if the first marble is green, since there is only 1 red marble in the bag. Outcome is \((G, R)\). \(P(G \text{ then } R) = \frac{8}{9} \times \frac{1}{8} = \frac{1}{9}\).

Common Mistakes

  • !Forgetting to reduce the denominator and numerator in consecutive selections without replacement, treating dependent events as independent.
  • !Assuming that events are dependent simply because they involve two stages, even when items are returned with replacement.
  • !Adding probabilities instead of multiplying them when determining the joint probability of combined events occurring simultaneously or consecutively, \(P(A \cap B)\).

Misconceptions

Misconception

Independent events always occur at the same time.

Correct Fact

Events can be independent even if they occur consecutively, provided the outcome of the first has no effect on the outcome of the second, such as rolling a die twice or drawing cards with replacement.

Misconception

The multiplication rule \(P(A \text{ and } B) = P(A) \times P(B)\) can be directly applied with original individual probabilities to dependent events without adjusting the probability of \(B\).

Correct Fact

For dependent events, the probability of the second event depends on the outcome of the first event. Conditional adjustments, such as those shown on tree diagram branches, must be accounted for before multiplying.

Summary

  • Independent events: Occurrence of event \(A\) has no effect on the occurrence of event \(B\).
  • Dependent events: Occurrence of event \(A\) affects the occurrence of event \(B\).
  • Multiplication rule for independent events: \(P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)\).
  • Methods to solve: listing sample space outcomes using \(P(E) = \frac{n(E)}{n(S)}\), the multiplication rule, and tree diagrams with probability branches.

Short Activity

Complete all parts first, then press the check answer button.

A. Objective Quiz

1 A fair coin is tossed and a fair six-faced die is rolled. Which statement best classifies these combined events?

2 Box \(P\) contains 2 red cards and 3 black cards. Box \(Q\) contains 5 red cards and 6 green cards. A card is drawn from Box \(P\) and placed into Box \(Q\), and then a card is drawn from Box \(Q\). The event of drawing a card from Box \(Q\) is:

B. Fill in the Blanks

3 The multiplication rule of probability states that for two independent events \(A\) and \(B\), \(P(A \text{ and } B) = P(A) \times\) .

4 In a probability tree diagram, each represents a possible outcome of an event.

C. Drag and Drop: Match Each Scenario

Drag each scenario to the correct classification or probability concept. On a phone, tap the answer choice first, then tap the matching box.

Scenario
Tossing a fair coin twice consecutively
Selecting two bulbs one by one from a box without replacement
Formula \(P(A \cap B \cap C) = P(A) \times P(B) \times P(C)\)
Diagram that displays all possible outcomes where each branch represents an outcome
Classification / Concept
Independent eventsCorrectTry againEmpty
Dependent eventsCorrectTry againEmpty
Multiplication rule for three independent eventsCorrectTry againEmpty
Tree diagramCorrectTry againEmpty
 

Key Terms

Combined events Independent events Dependent events Without replacement With replacement Multiplication rule of probability Tree diagram