UK K12 GCSE/A-Level: Year 11 GCSE Mathematics - Probability, Tree Diagrams, Conditional Probability

Learning Objectives

By the end of this topic, students will be able to:

• Use tree diagrams to represent and calculate probabilities of independent and dependent events
• Apply the concept of conditional probability to solve problems involving tree diagrams
• Understand the difference between independent and dependent events and how to identify them
• Use probability rules to calculate probabilities of events and solve problems involving tree diagrams
• Apply probability to real-world scenarios and evaluate the reasonableness of calculated probabilities

Core Concepts

Probability is a measure of the likelihood of an event occurring. A probability is a number between 0 and 1, where 0 represents an impossible event and 1 represents a certain event. When two events are independent, the probability of both events occurring is the product of their individual probabilities. When two events are dependent, the probability of one event occurring affects the probability of the other event occurring.

A tree diagram is a visual representation of a sequence of events, where each event is represented by a branch. The probability of each event is represented by a number on the branch. Tree diagrams can be used to calculate probabilities of independent and dependent events.

Conditional probability is the probability of an event occurring given that another event has occurred. This is represented by the formula P(A|B) = P(A and B) / P(B), where P(A|B) is the probability of event A occurring given that event B has occurred.

Worked Examples

Example 1: Independent Events

Tom has 5 red balls and 3 blue balls in a bag. He draws one ball at random. What is the probability that he draws a red ball and then a blue ball?

We can use a tree diagram to represent this situation:

          +---------------+
          |  Red (5/8)    |
          +---------------+
                  |
                  |
                  v
          +---------------+
          |  Blue (3/7)    |
          +---------------+

The probability of drawing a red ball is 5/8, and the probability of drawing a blue ball given that a red ball has been drawn is 3/7. Since the events are independent, we can multiply the probabilities to get the probability of both events occurring:

(5/8) × (3/7) = 15/56

Example 2: Dependent Events

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is red and the second ball drawn is blue?

We can use a tree diagram to represent this situation:

          +---------------+
          |  Red (5/8)    |
          +---------------+
                  |
                  |
                  v
          +---------------+
          |  Blue (3/7)    |
          +---------------+

However, since the second ball is drawn without replacing the first ball, the probability of drawing a blue ball given that a red ball has been drawn is not 3/7. Instead, it is 3/7 × (5/8) = 15/56, since there are now only 5 red balls and 3 blue balls left in the bag.

The probability of drawing a red ball and then a blue ball is:

(5/8) × (15/56) = 75/448

Example 3: Conditional Probability

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and it is found to be red. What is the probability that the ball drawn is blue?

We can use the formula for conditional probability:

P(Blue|Red) = P(Blue and Red) / P(Red)

Since the events are dependent, we need to use the formula for dependent events:

P(Blue and Red) = P(Red) × P(Blue|Red)

P(Red) = 5/8 P(Blue|Red) = 3/7

P(Blue and Red) = (5/8) × (3/7) = 15/56

P(Red) = 5/8

P(Blue|Red) = (15/56) / (5/8) = 3/14

Common Misconceptions

• Many students assume that tree diagrams only apply to independent events. However, tree diagrams can also be used to represent dependent events.
• Some students mistakenly assume that conditional probability only applies to dependent events. However, conditional probability can be used to solve problems involving both independent and dependent events.
• Students may struggle to identify the difference between independent and dependent events. To overcome this, it is essential to practice identifying and solving problems involving both types of events.

Exam Tips

• When using tree diagrams, make sure to label each branch with the probability of the event occurring.
• When solving problems involving conditional probability, make sure to use the correct formula and to identify the given and unknown probabilities.
• When solving problems involving dependent events, make sure to take into account the effect of the first event on the probability of the second event.
• When evaluating the reasonableness of calculated probabilities, make sure to consider the context of the problem and the given information.

MCQs with Explanations

MCQ 1 [F]

What is the probability of drawing a red ball and then a blue ball from a bag containing 5 red balls and 3 blue balls?

A) 15/56 B) 25/56 C) 35/56 D) 45/56

Correct answer: A) 15/56

Why the distractors fail: * B) 25/56 is incorrect because it assumes that the events are independent, but they are actually dependent. * C) 35/56 is incorrect because it assumes that the probability of drawing a red ball is 5/7, but it is actually 5/8. * D) 45/56 is incorrect because it assumes that the probability of drawing a blue ball given that a red ball has been drawn is 3/7, but it is actually 15/56.

MCQ 2 [H]

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is red and the second ball drawn is blue?

A) 15/56 B) 25/56 C) 35/56 D) 45/56

Correct answer: A) 15/56

Why the distractors fail: * B) 25/56 is incorrect because it assumes that the events are independent, but they are actually dependent. * C) 35/56 is incorrect because it assumes that the probability of drawing a blue ball given that a red ball has been drawn is 3/7, but it is actually 15/56. * D) 45/56 is incorrect because it assumes that the probability of drawing a red ball is 5/7, but it is actually 5/8.

MCQ 3 [F]

What is the probability of drawing a blue ball given that a red ball has been drawn from a bag containing 5 red balls and 3 blue balls?

A) 3/14 B) 3/7 C) 5/14 D) 5/7

Correct answer: A) 3/14

Why the distractors fail: * B) 3/7 is incorrect because it assumes that the events are independent, but they are actually dependent. * C) 5/14 is incorrect because it assumes that the probability of drawing a blue ball given that a red ball has been drawn is 3/7, but it is actually 3/14. * D) 5/7 is incorrect because it assumes that the probability of drawing a blue ball is 3/7, but it is actually 3/14.

MCQ 4 [H]

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and it is found to be red. What is the probability that the ball drawn is blue?

A) 3/14 B) 3/7 C) 5/14 D) 5/7

Correct answer: A) 3/14

Why the distractors fail: * B) 3/7 is incorrect because it assumes that the events are independent, but they are actually dependent. * C) 5/14 is incorrect because it assumes that the probability of drawing a blue ball given that a red ball has been drawn is 3/7, but it is actually 3/14. * D) 5/7 is incorrect because it assumes that the probability of drawing a blue ball is 3/7, but it is actually 3/14.

MCQ 5 [H]

A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is blue and the second ball drawn is red?

A) 15/56 B) 25/56 C) 35/56 D) 45/56

Correct answer: C) 35/56

Why the distractors fail: * A) 15/56 is incorrect because it assumes that the events are independent, but they are actually dependent. * B) 25/56 is incorrect because it assumes that the probability of drawing a blue ball given that a red ball has been drawn is 3/7, but it is actually 15/56. * D) 45/56 is incorrect because it assumes that the probability of drawing a red ball is 5/7, but it is actually 5/8.

Short-answer questions

1. A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is red and the second ball drawn is blue?

(Answer should be 15/56)

1. A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and it is found to be red. What is the probability that the ball drawn is blue?

(Answer should be 3/14)

1. A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is blue and the second ball drawn is red?

(Answer should be 35/56)

1. A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and it is found to be blue. What is the probability that the ball drawn is red?

(Answer should be 5/14)

1. A bag contains 5 red balls and 3 blue balls. A ball is drawn at random, and then a second ball is drawn without replacing the first ball. What is the probability that the first ball drawn is red and the second ball drawn is red?

(Answer should be 25/56)