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Class 12 Science (PCM)

Mathematics Part 2

Probability

Probability

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Binomial Probability Applications
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Bayes’ Theorem Word Problems
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Probability Distribution of a Random Variable
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Three Mutually Independent Events
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Theorem of Total Probability
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Multiplication Rule of Probability
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High
Reverse Probability and Causes
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High
Bayes' Theorem
Accuracy: 38%
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Conditional Probability Properties
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Conditional Probability with Reduced Sample Space
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Independent vs Mutually Exclusive Events
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Prior and Posterior Probability
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Probability Without Replacement
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Independent vs Dependent Events
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Multiplication Rule for Three or More Events
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Conditional Probability
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Independent Events
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Bernoulli Trials
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Partition of Sample Space
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Learning Objectives

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Learning Objectives

  • Understand the concept of probability as a measure of uncertainty in random experiments.
  • Explain the axiomatic approach to probability as formulated by A.N. Kolmogorov.
  • Apply the addition rule of probability to calculate probabilities of events.
  • Define and calculate conditional probability of an event given another event has occurred.
  • Utilize Bayes' theorem to find reverse probabilities in various scenarios.
  • Recognize the concept of independence of events and apply it to solve problems.
  • Define random variables and their probability distributions, including mean and variance.
  • Analyze and apply the binomial distribution in discrete probability scenarios.

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Revision Notes & Summary

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Chapter 13: Probability

13.1 Introduction

  • Probability as a measure of uncertainty in random experiments.
  • Axiomatic approach by A.N. Kolmogorov.
  • Equivalence between axiomatic and classical theories for equally likely outcomes.
  • Topics covered:
    • Conditional probability
    • Bayes' theorem
    • Multiplication rule of probability
    • Independence of events
    • Random variables and their distributions
    • Mean and variance of probability distributions
    • Binomial distribution

13.2 Conditional Probability

  • Definition: The probability of an event given that another event has occurred.
  • Example: Tossing three fair coins.
    • Sample space: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Key Concepts

  • Conditional Probability Formula:
    P(E|F) = P(E ∩ F) / P(F), P(F) ≠ 0
  • Independence: If E and F are independent, then:
    • P(E ∩ F) = P(E) * P(F)
    • P(E|F) = P(E)
    • P(F|E) = P(F)

Theorem of Total Probability

  • For a partition {E₁, E₂, ..., Eₙ} of a sample space:
    • P(A) = Σ P(Eᵢ) * P(A|Eᵢ)

Bayes' Theorem

  • For events E₁, E₂, ..., Eₙ that partition S:
    • P(Eᵢ|A) = [P(Eᵢ) * P(A|Eᵢ)] / Σ [P(Eⱼ) * P(A|Eⱼ)]

Examples

  1. Example of Conditional Probability:
    • If a machine is correctly set up, it produces 90% acceptable items.
    • If incorrectly set up, it produces 40% acceptable items.
    • Given that 80% of setups are correct, find the probability of correct setup given 2 acceptable items produced.
  2. Example of Bayes' Theorem:
    • Bag I: 3 red, 4 black balls; Bag II: 5 red, 6 black balls.
    • Find the probability that a red ball drawn came from Bag II.

Exercises

  1. Calculate conditional probabilities based on various scenarios involving events A and B.
  2. Apply Bayes' theorem to real-world problems involving medical tests and card draws.
  3. Explore the implications of independence in probability events.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

0.024

B.

0.06

C.

0.08

D.

0.12
Correct Answer: A

Solution:

Since AA, BB, and CC are mutually independent, P(A∩B∩C)=P(A)⋅P(B)⋅P(C)P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C). We have P(A∩B)=P(A)⋅P(B)=0.2P(A \cap B) = P(A) \cdot P(B) = 0.2, P(B∩C)=P(B)⋅P(C)=0.3P(B \cap C) = P(B) \cdot P(C) = 0.3, and P(A∩C)=P(A)⋅P(C)=0.4P(A \cap C) = P(A) \cdot P(C) = 0.4. Solving these equations gives P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4, P(C)=0.6P(C) = 0.6. Therefore, P(A∩B∩C)=0.5⋅0.4⋅0.6=0.12P(A \cap B \cap C) = 0.5 \cdot 0.4 \cdot 0.6 = 0.12.

Chapter Concept:

Three Mutually Independent Events

A.

It varies with each trial.

B.

It remains constant across trials.

C.

It depends on the outcome of the previous trial.

D.

It is always 0.5.
Correct Answer: B

Solution:

In Bernoulli trials, the probability of success remains constant across trials.

Chapter Concept:

Bernoulli Trials

A.

0.15

B.

0.20

C.

0.30

D.

0.50
Correct Answer: A

Solution:

For independent events, P(E∩F)=P(E)⋅P(F)=0.5×0.3=0.15P(E \cap F) = P(E) \cdot P(F) = 0.5 \times 0.3 = 0.15.

Chapter Concept:

Independent Events

A.

12\frac{1}{2}

B.

13\frac{1}{3}

C.

38\frac{3}{8}

D.

58\frac{5}{8}
Correct Answer: C

Solution:

Using Bayes' theorem, the probability that it actually was a six given the person reports a six is P(S1∣E)=P(S1)P(E∣S1)P(S1)P(E∣S1)+P(S2)P(E∣S2)=16×3416×34+56×14=38P(S_1|E) = \frac{P(S_1) P(E|S_1)}{P(S_1) P(E|S_1) + P(S_2) P(E|S_2)} = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{8}.

Chapter Concept:

Reverse Probability and Causes

A.

0.35

B.

0.38

C.

0.42

D.

0.46
Correct Answer: B

Solution:

According to the theorem of total probability, P(A)=P(E1)P(A∣E1)+P(E2)P(A∣E2)+P(E3)P(A∣E3)=0.2×0.1+0.5×0.4+0.3×0.6=0.38P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3) = 0.2 \times 0.1 + 0.5 \times 0.4 + 0.3 \times 0.6 = 0.38.

Chapter Concept:

Theorem of Total Probability

A.

P(A∩B)=P(A)+P(B)P(A \cap B) = P(A) + P(B)

B.

P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B)

C.

P(A∩B)=P(A)−P(B)P(A \cap B) = P(A) - P(B)

D.

P(A∩B)=P(B)−P(A)P(A \cap B) = P(B) - P(A)
Correct Answer: B

Solution:

For independent events, the probability of their intersection is the product of their individual probabilities: P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B).

Chapter Concept:

Independent vs Mutually Exclusive Events

A.

P(E)=0.5P(E) = 0.5

B.

P(E)=0.8P(E) = 0.8

C.

P(E)=1.0P(E) = 1.0

D.

P(E)=0.4P(E) = 0.4
Correct Answer: B

Solution:

Using the formula for conditional probability, P(E∣F)=P(E∩F)P(F)P(E|F) = \frac{P(E \cap F)}{P(F)}, we can solve for P(E)P(E): P(E)=P(E∩F)P(E∣F)=0.20.5=0.4P(E) = \frac{P(E \cap F)}{P(E|F)} = \frac{0.2}{0.5} = 0.4.

Chapter Concept:

Conditional Probability Properties

A.

0.5

B.

0.3

C.

0.4

D.

0.2
Correct Answer: B

Solution:

Using Bayes' theorem, the probability that the doctor came by train given he is late is: P(T1∣E)=P(T1)P(E∣T1)P(T1)P(E∣T1)+P(T2)P(E∣T2)+P(T3)P(E∣T3)+P(T4)P(E∣T4)=0.3×0.40.3×0.4+0.5×0.3+0.1×0.1+0.1×0=0.3P(T_1|E) = \frac{P(T_1)P(E|T_1)}{P(T_1)P(E|T_1) + P(T_2)P(E|T_2) + P(T_3)P(E|T_3) + P(T_4)P(E|T_4)} = \frac{0.3 \times 0.4}{0.3 \times 0.4 + 0.5 \times 0.3 + 0.1 \times 0.1 + 0.1 \times 0} = 0.3.

Chapter Concept:

Bayes’ Theorem Word Problems

A.

13\frac{1}{3}

B.

25\frac{2}{5}

C.

14\frac{1}{4}

D.

314\frac{3}{14}
Correct Answer: D

Solution:

The probability of drawing a black ball first is 515=13\frac{5}{15} = \frac{1}{3}. After drawing a black ball, there are 10 white balls and 4 black balls left. The probability of then drawing a white ball is 1014=57\frac{10}{14} = \frac{5}{7}. Therefore, the probability of this sequence is 13×57=521\frac{1}{3} \times \frac{5}{7} = \frac{5}{21}.

Chapter Concept:

Probability Without Replacement

A.

Yes, because P(E∣F)=P(E)P(E|F) = P(E)

B.

No, because P(E∩F)≠P(E)⋅P(F)P(E \cap F) \neq P(E) \cdot P(F)

C.

Yes, because P(F∣E)=P(F)P(F|E) = P(F)

D.

No, because P(E∣F)≠P(E)P(E|F) \neq P(E)
Correct Answer: D

Solution:

Given P(E)=12P(E) = \frac{1}{2}, P(F)=12P(F) = \frac{1}{2}, and P(E∣F)=14P(E|F) = \frac{1}{4}, P(E∣F)≠P(E)P(E|F) \neq P(E), hence EE and FF are not independent.

Chapter Concept:

Independent Events

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True or False

Correct Answer: True

Solution:

The multiplication rule for three events is correctly stated as P(E∩F∩G)=P(E)P(F∣E)P(G∣E∩F)P(E \cap F \cap G) = P(E) P(F|E) P(G|E \cap F), which extends to four or more events.

Chapter Concept :

Multiplication Rule for Three or More Events

Correct Answer: True

Solution:

A partition of a sample space SS is defined by events that are pairwise disjoint, exhaustive (i.e., their union is the entire sample space SS), and each event has a non-zero probability. This ensures that all possible outcomes are covered without overlap.

Chapter Concept :

Partition of Sample Space

Correct Answer: True

Solution:

By definition, the conditional probability P(E∣F)P(E|F) satisfies the property 0≤P(E∣F)≤10 \leq P(E|F) \leq 1.

Chapter Concept :

Conditional Probability Properties

Correct Answer: True

Solution:

In sequential draws without replacement, the probability of subsequent events changes based on the outcomes of previous events. This is a classic example of conditional probability where the sample space is reduced after each draw.

Chapter Concept :

Probability Without Replacement

Correct Answer: True

Solution:

The multiplication rule for three events is correctly stated as P(E∩F∩G)=P(E)⋅P(F∣E)⋅P(G∣E∩F)P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|E \cap F), which follows from the definition of conditional probability and the multiplication rule for two events.

Chapter Concept :

Multiplication Rule for Three or More Events

Correct Answer: True

Solution:

The multiplication rule of probability is correctly stated as P(E∩F)=P(E)P(F∣E)=P(F)P(E∣F)P(E \cap F) = P(E)P(F|E) = P(F)P(E|F), where P(F∣E)P(F|E) and P(E∣F)P(E|F) are the conditional probabilities of FF given EE and EE given FF, respectively.

Chapter Concept :

Multiplication Rule of Probability

Correct Answer: True

Solution:

The formula for conditional probability is correctly given as P(E∣F)=P(E∩F)P(F)P(E|F) = \frac{P(E \cap F)}{P(F)}, where P(F)≠0P(F) \neq 0 is a necessary condition to ensure the denominator is not zero.

Chapter Concept :

Conditional Probability

Correct Answer: True

Solution:

When drawing without replacement, the sample space changes after each draw, affecting the conditional probabilities. For instance, the probability of drawing a second black ball depends on the first draw, as the total number of balls and the number of black balls remaining have changed.

Chapter Concept :

Probability Without Replacement

Correct Answer: True

Solution:

The multiplication rule of probability is correctly stated as P(E∩F)=P(E)P(F∣E)=P(F)P(E∣F)P(E \cap F) = P(E)P(F|E) = P(F)P(E|F), and it can indeed be extended to more than two events, as shown in the provided excerpts.

Chapter Concept :

Multiplication Rule of Probability

Correct Answer: True

Solution:

By definition, two events EE and FF are independent if the probability of their intersection is equal to the product of their individual probabilities, i.e., P(E∩F)=P(E)⋅P(F)P(E \cap F) = P(E) \cdot P(F). This is a fundamental property of independent events.

Chapter Concept :

Independent Events

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