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Chapter Analysis
Intermediate28 pages • EnglishQuick Summary
The chapter on Probability discusses fundamental concepts related to probability such as conditional probability, independent events, and Bayes' theorem. It includes various examples to illustrate the multiplication rule of probability and concepts like mutually exclusive and exhaustive events. Additionally, the chapter covers the theorem of total probability and practical applications of probability in different scenarios.
Key Topics
- •Conditional probability
- •Independent events
- •Multiplication rule of probability
- •Bayes' Theorem
- •Mutually exclusive events
- •Theorem of total probability
- •Probability distributions
- •Random variables
Learning Objectives
- ✓Understand and calculate conditional probabilities
- ✓Identify and evaluate independent events
- ✓Apply Bayes’ theorem in real-world scenarios
- ✓Use the multiplication rule for calculating joint probabilities
- ✓Distinguish between mutually exclusive and independent events
- ✓Solve problems using the theorem of total probability
Questions in Chapter
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P (F|E)
Page 414
If P(A) = 3/5 and P(B) = 1/5, find P (A ∩ B) if A and B are independent events.
Page 422
A and B are two events such that P (A) ≠ 0. Find P(B|A), if (i) A is a subset of B (ii) A ∩ B = φ
Page 435
Additional Practice Questions
What is the probability that the sum of the numbers appearing on two thrown dice is greater than 8?
mediumAnswer: For the sum to be greater than 8, the following combinations are possible: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6). Total outcomes = 36, favorable outcomes = 10. Probability = 10/36 = 5/18.
Suppose you draw two cards from a deck without replacement. What is the probability that they are both aces?
mediumAnswer: Probability of first card being an ace = 4/52. Probability of second card being an ace = 3/51 (since one ace is already drawn). Therefore, P(both aces) = (4/52) * (3/51) = 1/221.
In a group of students, 40% like mathematics, 30% like physics, and 10% like both subjects. What is the probability that a student likes mathematics or physics?
easyAnswer: Using the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B), P(mathematics or physics) = 0.40 + 0.30 - 0.10 = 0.60.
A die is rolled three times. What is the probability of getting a score of six at least once?
mediumAnswer: Probability of not getting a six in one roll = 5/6. Probability of not getting a six in three rolls = (5/6)^3. Therefore, probability of getting at least one six = 1 - (5/6)^3 = 91/216.
In a classroom, 70% of the students passed in math, 80% passed in science, and 60% passed in both. What is the probability that a student fails in both subjects?
easyAnswer: Probability of passing in at least one subject = P(math) + P(science) - P(both) = 0.70 + 0.80 - 0.60 = 0.90. Probability of failing both = 1 - 0.90 = 0.10.