If $P(A|B) > P(A)$, which of the following is true?

Correct Option: C. Solution: If $P(A|B) > P(A)$, it implies that $B$ increases the likelihood of $A$, hence $P(B|A) > P(B)$.

Which of the following statements is true if $P(A|B) > P(A)$?

Correct Option: C. Solution: If $P(A|B) > P(A)$, it implies that knowing $B$ increases the likelihood of $A$, so $P(B|A) > P(B)$.

If $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \cap B) = 0.1$, what is the probability of $A$ given $B$, $P(A|B)$?

Correct Option: C. Solution: By definition, $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.3} = 0.333$.

Two events $A$ and $B$ are independent. If $P(A) = 0.5$ and $P(B) = 0.6$, what is $P(A \cap B)$?

Correct Option: B. Solution: For independent events, $P(A \cap B) = P(A)P(B) = 0.5 \times 0.6 = 0.3$.

In a random experiment, if $P(A) = 0.3$ and $P(B) = 0.6$, what is $P(A \cup B)$ if $A$ and $B$ are mutually exclusive?

Correct Option: A. Solution: For mutually exclusive events, $P(A \cup B) = P(A) + P(B) = 0.3 + 0.6 = 0.9$.

Probability

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Summary

Chapter Summary: Probability

Key Concepts

Probability Range: 0 ≤ P(E) ≤ 1
Conditional Probability:
- P(E|F) = P(E ∩ F) / P(F) (P(F) ≠ 0)
- P(E'|F) = 1 - P(E|F)
Addition Rule: P(E ∪ F | G) = P(E | G) + P(F | G) - P(E ∩ F | G)
Multiplication Rule:
- P(E ∩ F) = P(E) P(F|E) (P(E) ≠ 0)
- P(E ∩ F) = P(F) P(E|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 S:
- P(A) = Σ P(Eᵢ) P(A|Eᵢ) for i = 1 to n

Bayes' Theorem

For events E₁, E₂, ..., Eₙ that partition S:
- P(Eᵢ | A) = [P(Eᵢ) P(A|Eᵢ)] / Σ [P(Eⱼ) P(A|Eⱼ)] for j = 1 to n

Examples

Example of Conditional Probability: If a student is known to have an A grade, what is the probability they are a hostler?
Example of Bayes' Theorem: Given a red ball drawn from a bag, find the probability it came from a specific bag.

Important Formulas

Conditional Probability: P(A|B) = P(A ∩ B) / P(B)
Total Probability: P(A) = Σ P(Eᵢ) P(A|Eᵢ)
Bayes' Theorem: P(Eᵢ | A) = [P(Eᵢ) P(A|Eᵢ)] / Σ [P(Eⱼ) P(A|Eⱼ)]

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.

Detailed Notes

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

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.
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

Calculate conditional probabilities based on various scenarios involving events A and B.
Apply Bayes' theorem to real-world problems involving medical tests and card draws.
Explore the implications of independence in probability events.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Probability

Common Pitfalls

Misunderstanding Conditional Probability: Students often confuse conditional probability with joint probability. Remember that P(A|B) is not the same as P(A and B).
Ignoring Independence: When events are independent, P(A|B) = P(A). Failing to recognize this can lead to incorrect calculations.
Incorrect Application of Bayes' Theorem: Ensure you correctly identify the prior probabilities and the likelihoods when applying Bayes' theorem.
Assuming Events are Mutually Exclusive: Not all events are mutually exclusive. Be cautious when applying formulas that assume exclusivity.

Tips for Success

Practice with Examples: Work through examples that involve conditional probability and Bayes' theorem to solidify your understanding.
Draw Sample Spaces: For complex problems, drawing a sample space can help visualize the events and their relationships.
Review Key Formulas: Familiarize yourself with key formulas such as the multiplication rule and the theorem of total probability.
Check Your Work: Always double-check calculations, especially when dealing with fractions or percentages, to avoid simple arithmetic errors.

Practice & Assessment

Multiple Choice Questions

0.10

0.25

0.50

0.35

Correct Answer: B

Solution:

Let

D

be the event of producing a defective item,

A

B

, and

C

be the events of the item being produced by operators A, B, and C respectively. We have

P(D|A) = 0.01

P(D|B) = 0.05

P(D|C) = 0.07

P(A) = 0.5

P(B) = 0.3

P(C) = 0.2

. Using Bayes' theorem:

P(A|D) = \frac{P(A)P(D|A)}{P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C)} = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = 0.25.

0.80

0.90

0.95

0.85

Correct Answer: C

Solution:

Let

A

be the event that the machine produces 2 acceptable items,

B_1

be the event of correct setup, and

B_2

be the event of incorrect setup. We have

P(B_1) = 0.8

P(B_2) = 0.2

P(A|B_1) = 0.9 \times 0.9

, and

P(A|B_2) = 0.4 \times 0.4

. Using Bayes' theorem:

P(B_1|A) = \frac{P(B_1) P(A|B_1)}{P(B_1) P(A|B_1) + P(B_2) P(A|B_2)} = \frac{0.8 \times 0.81}{0.8 \times 0.81 + 0.2 \times 0.16} = 0.95.

0.4096

0.7373

0.2048

0.8192

Correct Answer: A

Solution:

This is a binomial probability problem where

n = 5

p = 0.8

, and

k = 4

. The probability is given by

P(X = 4) = \binom{5}{4} (0.8)^4 (0.2)^1 = 5 \times 0.4096 \times 0.2 = 0.4096

\frac{1}{6}

\frac{1}{36}

\frac{1}{216}

\frac{1}{3}

Correct Answer: A

Solution:

The probability of a 4 on the third toss is

\frac{1}{6}

, independent of the previous outcomes.

P(B|A) < P(B)

P(A \cap B) < P(A) \cdot P(B)

P(B|A) > P(B)

P(B|A) = P(B)

Correct Answer: C

Solution:

P(A|B) > P(A)

, it implies that

B

increases the likelihood of

A

, hence

P(B|A) > P(B)

1/2

1/3

2/3

1/4

Correct Answer: C

Solution:

Let

D_1

be the event of throwing 1, 2, 3, or 4, and

D_2

be the event of throwing 5 or 6. The probability of getting exactly one head if

D_1

occurs is 1/2 (since she tosses once). The probability of getting exactly one head if

D_2

occurs is 3/8 (from binomial distribution with

n=3

and

p=1/2

). Using Bayes' theorem,

P(D_1|H) = \frac{P(H|D_1)P(D_1)}{P(H|D_1)P(D_1) + P(H|D_2)P(D_2)} = \frac{(1/2)(2/3)}{(1/2)(2/3) + (3/8)(1/3)} = \frac{2/6}{2/6 + 1/8} = \frac{2/3}.

0.0

0.35

0.5

1.2

Correct Answer: A

Solution:

For mutually exclusive events,

P(A \cap B) = 0

because they cannot occur simultaneously.

1/3

1/4

1/2

2/3

Correct Answer: A

Solution:

Let

B_1

be the event that at least one child is a boy and

B_2

be the event that both children are boys. The sample space for two children is:

\{(B,B), (B,G), (G,B), (G,G)\}

. Given at least one is a boy, the possible outcomes are

\{(B,B), (B,G), (G,B)\}

. Thus,

P(B_2|B_1) = \frac{1}{3}

\frac{2}{3}

\frac{1}{3}

\frac{1}{2}

\frac{2}{5}

Correct Answer: A

Solution:

The conditional probability

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = \frac{2}{3}

Yes, they are independent.

No, they are not independent.

They are mutually exclusive.

They are both dependent and mutually exclusive.

Correct Answer: C

Solution:

Events

A

and

B

are mutually exclusive because they cannot occur simultaneously. If the sum is 7, it cannot be 11 at the same time.

\frac{1}{6}

\frac{1}{36}

\frac{1}{216}

\frac{1}{3}

Correct Answer: A

Solution:

The probability of getting a 4 on the third toss is independent of the previous results, so it remains

\frac{1}{6}

0.3

0.5

0.1

0.6

Correct Answer: A

Solution:

Since

A

and

B

are independent, the probability that both occur is given by

P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.5 = 0.3.

0.24

0.36

0.48

0.52

Correct Answer: C

Solution:

The probability of getting exactly one head in two tosses is given by the binomial probability:

P(X=1) = \binom{2}{1} (0.6)^1 (0.4)^1 = 2 \times 0.6 \times 0.4 = 0.48.

P(B|A) < P(B)

P(A \cap B) < P(A) \cdot P(B)

P(B|A) > P(B)

P(B|A) = P(B)

Correct Answer: C

Solution:

P(A|B) > P(A)

, it implies that knowing

B

increases the likelihood of

A

, so

P(B|A) > P(B)

Yes, because

P(A \cap B) = P(A) \times P(B)

No, because

P(A \cap B) \neq P(A) \times P(B)

Yes, because

P(A|B) = P(A)

No, because

P(B|A) \neq P(B)

Correct Answer: A

Solution:

For two events to be independent,

P(A \cap B) = P(A) \times P(B)

. Here,

0.2 = 0.5 \times 0.4

, which holds true. Therefore, A and B are independent.

\frac{4}{13}

\frac{17}{52}

\frac{1}{4}

\frac{5}{13}

Correct Answer: D

Solution:

There are 13 spades and 4 aces in a deck, but 1 card is both a spade and an ace. So,

P( ext{spade or ace}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}

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

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

P(A \cup B) = P(A) P(B)

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Correct Answer: B

Solution:

For independent events, the probability of both events occurring is the product of their individual probabilities, i.e.,

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

0.2

0.4

0.3

0.5

Correct Answer: C

Solution:

Using Bayes' theorem,

P(B|D) = \frac{P(B) \cdot P(D|B)}{P(D)} = \frac{0.4 \cdot 0.01}{0.6 \cdot 0.02 + 0.4 \cdot 0.01} = \frac{0.004}{0.012 + 0.004} = \frac{0.004}{0.016} = 0.25

1/6

1/3

1/2

2/3

Correct Answer: C

Solution:

There are three even numbers on a die (2, 4, 6). Thus, the probability is

\frac{3}{6} = \frac{1}{2}

0.2

0.25

0.3

0.5

Correct Answer: B

Solution:

The possible outcomes for the second roll to achieve a sum of 8 with the first roll being 3 are:

5

. The probability of rolling a 5 on a fair die is

\frac{1}{6}

. Therefore, the probability is

\frac{1}{6} \approx 0.25

Yes, they are independent.

No, they are not independent.

Cannot be determined.

They are mutually exclusive.

Correct Answer: B

Solution:

For

A

and

B

to be independent,

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

. We have

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.58

. Therefore,

P(A \cap B) = 0.3 + 0.4 - 0.58 = 0.12

. But

P(A)P(B) = 0.3 \times 0.4 = 0.12

. Hence,

A

and

B

are not independent.

1/6

1/36

1/3

1/2

Correct Answer: A

Solution:

Given that the first number is 3, the possible outcomes for the second number to make the sum 8 are only 5. Thus,

P(E|F) = \frac{1}{6}

1/4

1/2

1/3

2/3

Correct Answer: B

Solution:

Let

G_1

be the event that the youngest child is a girl, and

G_2

be the event that both children are girls. The sample space for two children is {BB, BG, GB, GG}. Given that the youngest is a girl, the possible outcomes are {GB, GG}. Thus, the probability that both are girls given the youngest is a girl is

P(G_2|G_1) = \frac{1}{2}

\frac{4}{9}

\frac{5}{9}

\frac{1}{2}

\frac{5}{10}

Correct Answer: B

Solution:

The probability of drawing a red ball from Bag II is

\frac{5}{9}

because there are 5 red balls out of a total of 9 balls.

0.2

0.3

0.333

0.5

Correct Answer: C

Solution:

By definition,

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.3} = 0.333

0.4

0.2

0.333

0.6

Correct Answer: C

Solution:

Using Bayes' Theorem:

P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{0.01 \times 0.4}{0.02 \times 0.6 + 0.01 \times 0.4} = 0.333

0.1

0.2

0.3

0.4

Correct Answer: B

Solution:

Using the formula

P(A \cup B) = P(A) + P(B) - P(A \cap B)

, we find

P(A \cap B) = 0.5 + 0.4 - 0.7 = 0.2

0.7

0.9

0.6

0.8

Correct Answer: D

Solution:

For independent events,

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - (0.5 \times 0.4) = 0.8

0.8

0.7

0.9

1.0

Correct Answer: B

Solution:

Since

E

and

F

are independent,

P(E \cup F) = P(E) + P(F) - P(E)P(F) = 0.6 + 0.5 - 0.6 \times 0.5 = 0.7

\frac{1}{8}

\frac{7}{8}

\frac{1}{2}

\frac{3}{4}

Correct Answer: B

Solution:

The probability of not getting an odd number in one roll is

\frac{1}{2}

. Thus, the probability of not getting an odd number in three rolls is

(\frac{1}{2})^3 = \frac{1}{8}

. Therefore, the probability of getting an odd number at least once is

1 - \frac{1}{8} = \frac{7}{8}

0.5

0.7

0.75

0.85

Correct Answer: C

Solution:

Let

D

be the event that an item is defective. We need to find

P(X|D)

. Using Bayes' theorem,

P(X|D) = \frac{P(X)P(D|X)}{P(X)P(D|X) + P(Y)P(D|Y)} = \frac{0.7 \times 0.05}{0.7 \times 0.05 + 0.3 \times 0.02} = \frac{0.035}{0.035 + 0.006} = \frac{0.035}{0.041} \approx 0.75.

0.4

0.5

0.2

0.8

Correct Answer: A

Solution:

The conditional probability

P(A|B)

is given by

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4

0.20

0.40

0.80

0.60

Correct Answer: B

Solution:

Using the principle of inclusion-exclusion:

P(H \cup E) = P(H) + P(E) - P(H \cap E) = 0.6 + 0.4 - 0.2 = 0.8

. Thus,

P(

neither

) = 1 - 0.8 = 0.2

0.18

0.9

0.3

0.6

Correct Answer: A

Solution:

Since

A

and

B

are independent,

P(A \cap B) = P(A) \cdot P(B) = 0.6 \cdot 0.3 = 0.18

\frac{1}{6}

\frac{1}{5}

\frac{1}{4}

\frac{1}{3}

Correct Answer: B

Solution:

If the first number is 3, the possible outcomes for the second throw to make the sum 6 are (3,3). Thus, the probability is

\frac{1}{6}

, but since the first number is fixed as 3, the conditional probability is

\frac{1}{5}

considering the remaining outcomes.

0.3

0.4

0.5

0.8

Correct Answer: B

Solution:

Using the formula

P(A|B) = \frac{P(A \cap B)}{P(B)}

, we find

P(A \cap B) = P(A|B) \cdot P(B) = 0.8 \cdot 0.5 = 0.4

0.2

0.4

0.5

0.6

Correct Answer: B

Solution:

Using the formula

P(A|B) = \frac{P(A \cap B)}{P(B)}

, where

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3

. Thus,

P(A|B) = \frac{0.3}{0.5} = 0.6

0.2

0.4

0.6

0.8

Correct Answer: A

Solution:

The probability of reading at least one newspaper is

P(H \cup E) = P(H) + P(E) - P(H \cap E) = 0.6 + 0.4 - 0.2 = 0.8

. Thus, the probability of reading neither is

1 - 0.8 = 0.2

0.1

0.3

0.5

0.6

Correct Answer: B

Solution:

For independent events,

P(A \cap B) = P(A)P(B) = 0.5 \times 0.6 = 0.3

0.5

0.1

0.2

0.3

Correct Answer: C

Solution:

The intersection

E \cap F = \{3\}

. Since there are 6 outcomes in a die roll,

P(E \cap F) = \frac{1}{6} \approx 0.1667 \approx 0.2

\frac{1}{6}

\frac{1}{3}

\frac{1}{2}

\frac{1}{4}

Correct Answer: A

Solution:

The possible outcomes for the second die to make the sum 7 are 4 (since 3 + 4 = 7). There are 6 possible outcomes for the second die, so the probability is

\frac{1}{6}

1/6

1/36

1/3

1/18

Correct Answer: A

Solution:

The possible outcomes for the second roll to make the sum 8 are (3,5). Therefore, the probability is 1/6.

0.3828

0.4410

0.6572

0.7582

Correct Answer: C

Solution:

This is a binomial probability problem where

n = 4

p = 0.3

. We need to find

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

. Calculating each:

P(X = 0) = 0.2401

P(X = 1) = 0.4116

P(X = 2) = 0.2646

. Summing these gives

0.2401 + 0.4116 + 0.2646 = 0.6572

0.9

0.18

0.3

0.6

Correct Answer: A

Solution:

For mutually exclusive events,

P(A \cup B) = P(A) + P(B) = 0.3 + 0.6 = 0.9

0.9

0.8

0.7

0.6

Correct Answer: B

Solution:

The probability that a student passed at least one subject is given by

P(M \cup E) = P(M) + P(E) - P(M \cap E) = 0.7 + 0.6 - 0.4 = 0.9.

0.1

0.3

0.4

0.6

Correct Answer: C

Solution:

The probability that the first ball is red is

\frac{3}{5}

. If the first ball is red, the probability that the second ball is also red is

\frac{2}{4}

. Therefore, the probability that both balls are red is

\frac{3}{5} \times \frac{2}{4} = \frac{3}{10} = 0.3.

1/6

1/36

1/216

1/3

Correct Answer: A

Solution:

The sample space for rolling a die three times is

6^3 = 216

. Event

B

has 6 outcomes:

(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)

. Event

A \cap B

has 1 outcome:

(6,5,4)

. Thus,

P(B) = \frac{6}{216} = \frac{1}{36}

and

P(A \cap B) = \frac{1}{216}

. Therefore,

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/216}{1/36} = \frac{1}{6}

0.8

0.9

1.0

0.7

Correct Answer: A

Solution:

Using the formula for the probability of the union of two events,

P(M \cup P) = P(M) + P(P) - P(M \cap P) = 0.7 + 0.6 - 0.5 = 0.8.

0.875

0.5

0.125

0.25

Correct Answer: A

Solution:

The probability of not getting an odd number in one toss is

\frac{3}{6} = 0.5

. Therefore, the probability of not getting an odd number in three tosses is

0.5^3 = 0.125

. Thus, the probability of getting an odd number at least once is

1 - 0.125 = 0.875

0.1

0.2

0.3

0.4

Correct Answer: B

Solution:

The event

E

is 'number 4 appears at least once' and

F

is 'sum is 6'.

E = \{(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)\}

and

F = \{(2,4), (4,2)\}

P(E \cap F) = \frac{2}{36}

and

P(E) = \frac{11}{36}

. Thus,

P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{2/36}{11/36} = \frac{2}{11} \approx 0.18

1/2

1/4

1/8

Correct Answer: A

Solution:

Given that event

F

has occurred, the first two tosses are heads. The probability of getting a head on the third toss (event

E

) is

1/2

since the tosses are independent.

3/10

1/5

3/5

1/3

Correct Answer: A

Solution:

The probability of drawing two red balls is given by

\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}

0.12

0.70

0.10

0.24

Correct Answer: A

Solution:

Since

A

and

B

are independent,

P(A \cap B) = P(A) \cdot P(B) = 0.3 \cdot 0.4 = 0.12

\frac{3}{10}

\frac{1}{5}

\frac{1}{2}

\frac{2}{5}

Correct Answer: A

Solution:

The probability that the first ball is red is \frac{3}{5}. If the first ball is red, the probability that the second ball is also red is \frac{2}{4} = \frac{1}{2}. Thus, the probability that both balls are red is \frac{3}{5} \times \frac{1}{2} = \frac{3}{10}.

\frac{1}{2}

\frac{1}{3}

\frac{2}{3}

\frac{1}{6}

Correct Answer: A

Solution:

The sample space for

F

\\{5, 6\\}

and the event

E \cap F

\\{6\\}

. Therefore,

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/6}{2/6} = \frac{1}{2}

0.5

0.4

0.2

0.1

Correct Answer: A

Solution:

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5

0.67

0.5

0.33

0.2

Correct Answer: A

Solution:

The conditional probability

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = 0.67

0.12

0.58

0.7

0.82

Correct Answer: B

Solution:

For independent events,

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - (0.3 \cdot 0.4) = 0.58

1/4

3/4

1/2

1/3

Correct Answer: B

Solution:

The possible outcomes are HH, HT, TH, TT. The probability of getting at least one head is

1 - P(TT) = 1 - 1/4 = 3/4

True or False

Correct Answer: True

Solution:

For any two events

A

and

B

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. This holds true for independent events as well.

Correct Answer: True

Solution:

The independence of two events

E

and

F

implies that

P(E \cap F) = P(E)P(F)

. Similarly,

E'

and

F

are independent if

P(E' \cap F) = P(E')P(F)

. Since

P(E' \cap F) = P(F) - P(E \cap F)

and

P(E') = 1 - P(E)

, substituting these into the independence condition shows that

E'

and

F

are independent.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, meaning

P(A \cap B) = 0

. For independent events,

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

. If

P(A \cap B) = 0

, then either

P(A)

P(B)

must be zero for the events to be independent, which contradicts the definition of mutually exclusive events with nonzero probabilities.

Correct Answer: True

Solution:

By definition, two events are independent if the occurrence of one event does not affect the probability of the other event. This is mathematically expressed as

P(F|E) = P(F)

and

P(E|F) = P(E)

Correct Answer: True

Solution:

Bayes' theorem allows us to update the probability of an event based on new evidence or information about another event.

Correct Answer: True

Solution:

This is the definition of conditional probability.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, hence their intersection is zero. For independent events, the probability of their intersection is the product of their individual probabilities. Thus, mutually exclusive events cannot be independent unless one of the events has zero probability.

Correct Answer: True

Solution:

Using Bayes' Theorem,

P(B_1|A) = \frac{P(B_1)P(A|B_1)}{P(B_1)P(A|B_1) + P(B_2)P(A|B_2)} = 0.95

Correct Answer: True

Solution:

For any two events

A

and

B

, the probability of their union is given by

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Since

A

and

B

are independent,

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

, thus the formula holds.

Correct Answer: False

Solution:

Mutually exclusive events cannot be independent if they have nonzero probabilities, as their intersection is zero.

Correct Answer: True

Solution:

In a binomial distribution, the trials are independent, meaning the outcome of one trial does not affect the outcome of another.

Correct Answer: True

Solution:

The axiomatic approach to probability, formulated by A.N. Kolmogorov, treats probability as a function of the outcomes of an experiment.

Correct Answer: False

Solution:

The possible outcomes for the second die to achieve a sum greater than 9 are 5 or 6. Thus, the probability is

\frac{2}{6} = \frac{1}{3}

, which is less than 0.5.

Correct Answer: True

Solution:

P(A \cup B) = P(A)

, it implies that the occurrence of

B

does not add any new outcomes to the event

A

, meaning

B

does not occur when

A

occurs. Therefore,

P(B|A) = 0

Correct Answer: True

Solution:

Conditional probability

P(E|F)

is defined as

\frac{P(E \cap F)}{P(F)}

, which requires

P(F) \neq 0

to avoid division by zero.

Correct Answer: True

Solution:

By definition, two events

A

and

B

are independent if and only if

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

. This is the mathematical condition for independence.

Correct Answer: False

Solution:

This statement is false. If

P(A|B) > P(A)

, it implies that the occurrence of

B

increases the likelihood of

A

. This does not necessarily mean that

P(B|A) < P(B)

; the relationship between

P(B|A)

and

P(B)

depends on other factors and cannot be directly inferred from

P(A|B) > P(A)

Correct Answer: True

Solution:

By definition, probability values range from 0 (impossible event) to 1 (certain event).

Correct Answer: True

Solution:

For independent events,

P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.4 = 0.12

Correct Answer: True

Solution:

A binomial distribution assumes that each trial is independent and the probability of success is constant across trials.

Correct Answer: True

Solution:

This is one of the definitions of independence: the occurrence of

B

does not affect the probability of

A

Correct Answer: True

Solution:

By definition, two events

A

and

B

are independent if the probability of their intersection is the product of their individual probabilities.

Correct Answer: True

Solution:

For a complete set of mutually exclusive and exhaustive events, the total probability must cover the entire sample space, thus summing to 1.

Correct Answer: False

Solution:

Mutually exclusive events cannot be independent if they have nonzero probabilities, as they cannot occur simultaneously.

Correct Answer: True

Solution:

For any event

E

in a sample space

S

, the probability of the event and its complement

E'

is given by

P(E) + P(E') = 1

. This is a fundamental property of probability, as the event and its complement cover the entire sample space.

Correct Answer: True

Solution:

For independent events

E

and

F

, the probability of their intersection is given by

P(E \cap F) = P(E) \cdot P(F)

Correct Answer: True

Solution:

The only even prime number is 2, which is not a possible outcome on a single die roll.

Correct Answer: True

Solution:

The only even prime number is 2, and since a die does not have 2 as a face, the probability is 0.

Correct Answer: False

Solution:

The independence of events is not determined by their individual probabilities being equal. Independence requires that

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

. Without this condition, we cannot conclude independence.

Correct Answer: True

Solution:

For independent events

E

and

F

P(E \cap F') = P(E) \cdot (1 - P(F)) = P(E) \cdot P(F')

Correct Answer: True

Solution:

By definition, two events

E

and

F

are independent if

P(E \cap F) = P(E) \cdot P(F)

. This means the probability of both events occurring is indeed the product of their individual probabilities.

Correct Answer: True

Solution:

The probability of any event

E

is defined such that

0 \leq P(E) \leq 1

. This is a fundamental property of probability.

Correct Answer: True

Solution:

A die has numbers 1 to 6. The only even prime number is 2, so the probability of rolling a 2 is 1/6.

Correct Answer: True

Solution:

The only even prime number is 2. Since a die has numbers from 1 to 6, and only one number is even and prime (2), the probability of rolling a 2 on each die is

\left(\frac{1}{6}\right)^2 = \frac{1}{36}

. However, since the question asks for an even prime number on each die, and there are no other even primes, the statement is true as the probability is effectively 0.

Correct Answer: True

Solution:

A standard die has numbers 1 through 6. The only even prime number is 2, which can be rolled on a die. Therefore, the probability is not zero. However, if the context is misunderstood as 'each die', then the statement could be misleading.

Correct Answer: True

Solution:

By definition, two events are independent if the probability of their intersection is the product of their individual probabilities.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, so the probability of their intersection is zero.

Correct Answer: True

Solution:

The excerpt mentions that 20% of the students read both Hindi and English newspapers.

Correct Answer: False

Solution:

For events

A

and

B

to be mutually exclusive,

P(A \cup B)

should equal

P(A) + P(B)

. Here,

P(A \cup B) = 0.8

, but

P(A) + P(B) = 1.0

. Therefore,

A

and

B

are not mutually exclusive.

Correct Answer: True

Solution:

By definition, two events

A

and

B

are independent if the probability of their intersection is the product of their individual probabilities, i.e.,

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

. This is a fundamental property of independent events.

Correct Answer: True

Solution:

The probability of selecting three good oranges from a box of 15 (12 good, 3 bad) is calculated as

\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{11}{13}

Correct Answer: False

Solution:

Mutually exclusive events are those that cannot occur simultaneously, meaning

P(A \cap B) = 0

. The probabilities of

A

and

B

alone do not determine if they are mutually exclusive.

Correct Answer: True

Solution:

By definition, a biased coin that shows heads 75% of the time has a probability of 0.75 for heads.

Correct Answer: True

Solution:

The sum of the probabilities of all possible outcomes in a sample space is always equal to one, as it encompasses all possible events that can occur.

Correct Answer: False

Solution:

A leap year has 366 days, which is 52 weeks and 2 extra days. The extra days can be any pair of consecutive days. For a leap year to have 53 Tuesdays, one of the extra days must be a Tuesday. The probability is

\frac{2}{7}

, which is less than 0.5.

Correct Answer: False

Solution:

In a binomial distribution, the trials are independent of each other.

Correct Answer: True

Solution:

For mutually exclusive events,

P(A \\cup B) = P(A) + P(B)

. Thus,

P(A \\cup B) = 0.3 + 0.4 = 0.7

Correct Answer: True

Solution:

According to the definition of independent events,

E

and

F

are independent if

P(E \cap F) = P(E)P(F)

. This is a fundamental property of independent events.

Correct Answer: True

Solution:

The probability of obtaining a sum greater than 9, given that the black die shows 5, is indeed a conditional probability because it depends on the outcome of the black die.

Correct Answer: False

Solution:

P(A \mid B) > P(A)

, it implies that the occurrence of

B

increases the likelihood of

A

. This does not necessarily mean

P(B \mid A) < P(B)

; rather, it suggests a dependence between

A

and

B

. The relationship between

P(B \mid A)

and

P(B)

is not directly inferred from

P(A \mid B) > P(A)

Correct Answer: True

Solution:

Bayes' theorem provides a way to update the probability of a hypothesis based on new evidence, effectively allowing us to calculate the probability of a cause given an observed event.

Correct Answer: True

Solution:

A leap year has 366 days, which is 52 weeks and 2 extra days. The extra days can be any of the 7 combinations of two consecutive days, two of which include Tuesday.

Correct Answer: True

Solution:

E

and

F

are independent, then

P(E|F) = P(E)

by the definition of independent events.

Correct Answer: True

Solution:

The formula for the probability of the union of two independent events

A

and

B

P(A \cup B) = 1 - P(A')P(B')

. This is derived from the fact that

P(A \cup B) = P(A) + P(B) - P(A \cap B)

and for independent events,

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

Correct Answer: True

Solution:

Given the sample space and the events,

P(A|B) = \frac{1}{6}

as there is only one favorable outcome for

A

given

B

has occurred.

Correct Answer: False

Solution:

For

A

and

B

to be independent,

P(A \cap B)

should equal

P(A) \cdot P(B)

. However, given

P(A \cup B) = 0.8

, we find

P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.4 - 0.8 = 0.2

. Since

0.2 \neq 0.6 \cdot 0.4 = 0.24

A

and

B

are not independent.

Correct Answer: True

Solution:

A.N. Kolmogorov is credited with formulating the axiomatic approach to probability.

Correct Answer: True

Solution:

P(A \cup B) = P(A)

, it implies that

A \subseteq B

, as all elements of

A

are also in

B

Correct Answer: False

Solution:

The excerpt states that if a machine is incorrectly set up, it produces only 40% acceptable items, not 90%.

Correct Answer: True

Solution:

When two dice are rolled, the combinations for a sum of 4 are (1,3), (2,2), and (3,1), totaling 3 outcomes. For a sum of 3, the combinations are (1,2) and (2,1), totaling 2 outcomes. Therefore,

P(4) > P(3)

Correct Answer: True

Solution:

For two independent events

E

and

F

, the probability of their intersection is given by

P(E \cap F) = P(E) \cdot P(F)

Correct Answer: True

Solution:

Two events

A

and

B

are independent if

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

. Here,

P(A \cap B) = 0.25

and

P(A) \cdot P(B) = 0.5 \cdot 0.5 = 0.25

, so

A

and

B

are independent.

Correct Answer: True

Solution:

Bayes' theorem provides a way to update the probability of an event based on new evidence or information.

Correct Answer: True

Solution:

P(A|B) > P(A)

, it implies that the occurrence of

B

increases the likelihood of

A

. This means

P(A \cap B) = P(A|B)P(B) > P(A)P(B)

, indicating that

A

and

B

are not independent.

Correct Answer: True

Solution:

A

and

B

are independent, then

P(A \cap B) = P(A)P(B) = 0.5 \times 0.5 = 0.25

Correct Answer: True

Solution:

For independent events, the probability of the union is calculated as

P(A \cup B) = P(A) + P(B) - P(A \cap B)

, and since

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

for independent events, the formula simplifies to

P(A \cup B) = P(A) + P(B) - P(A)P(B)

Correct Answer: False

Solution:

The possible outcomes for the red die are 1, 2, or 3. For each of these, the black die must show a number such that the sum is 8. The combinations are (2,6), (3,5), and (1,7) (which is not possible). Only two combinations are valid, so the probability is

\frac{2}{18} = \frac{1}{9}

, which is approximately 0.11, less than 0.2.

Correct Answer: True

Solution:

For mutually exclusive events,

P(A \cup B) = P(A) + P(B)

. Given

P(A) = 0.5

and

P(B) = 0.5

P(A \cup B) = 0.5 + 0.5 = 1

Probability

AI Learning Assistant

Summary

Chapter Summary: Probability

Key Concepts

Theorem of Total Probability

Bayes' Theorem

Examples

Important Formulas

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 13: Probability

13.1 Introduction

13.2 Conditional Probability

Key Concepts

Theorem of Total Probability

Bayes' Theorem

Examples

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Probability

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.A manufacturer has three machine operators A, B, and C. A produces 1% defective items, B produces 5%, and C produces 7% defective items. A is on the job 50% of the time, B 30%, and C 20%. A defective item is produced. What is the probability it was produced by A?

Correct Answer: B

Q2.Consider a scenario where a machine is set up correctly 80% of the time. If set up correctly, it produces 90% acceptable items, and if set up incorrectly, it produces 40% acceptable items. If a machine produces 2 acceptable items in a row, what is the probability that it was set up correctly?

Correct Answer: C

Q3.A machine produces 80% of its products as non-defective. If 5 products are randomly selected, what is the probability that exactly 4 of them are non-defective?

Correct Answer: A

Q4.A die is thrown three times. What is the probability that a 4 appears on the third toss, given that a 6 and a 5 appeared on the first two tosses?

Correct Answer: A

Q5.If P(A∣B)>P(A)P(A|B) > P(A)P(A∣B)>P(A), which of the following is true?

Correct Answer: C

Q6.Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3, or 4, she tosses a coin once. If she obtained exactly one head, what is the probability that she threw 1, 2, 3, or 4 with the die?

Correct Answer: C

Q7.In a random experiment, if P(A)=0.7P(A) = 0.7P(A)=0.7 and P(B)=0.5P(B) = 0.5P(B)=0.5, and AAA and BBB are mutually exclusive, what is P(A∩B)P(A \cap B)P(A∩B)?

Correct Answer: A

Q8.A family has two children. What is the probability that both are boys given that at least one is a boy?

Correct Answer: A

Q9.In a random experiment, the probability of event AAA is 0.6 and the probability of event BBB is 0.3. If P(A∩B)=0.2P(A \cap B) = 0.2P(A∩B)=0.2, what is P(A∣B)P(A|B)P(A∣B)?

Correct Answer: A

Q10.Consider a scenario where two dice are rolled. Define event AAA as obtaining a sum of 7 and event BBB as getting a sum of 11. Are events AAA and BBB independent?

Correct Answer: C

Q11.A die is thrown three times. What is the probability of getting a 4 on the third toss given that a 6 and a 5 appeared on the first two tosses?

Correct Answer: A

Q12.If the probability of event AAA occurring is 0.6 and the probability of event BBB occurring is 0.5, and AAA and BBB are independent events, what is the probability that both AAA and BBB occur?

Correct Answer: A

Q13.A biased coin is such that the probability of getting a head is 0.6. If the coin is tossed twice, what is the probability of getting exactly one head?

Correct Answer: C

Q14.Which of the following statements is true if P(A∣B)>P(A)P(A|B) > P(A)P(A∣B)>P(A)?

Correct Answer: C

Q15.If two events A and B are such that P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.4P(B) = 0.4P(B)=0.4, and P(A∩B)=0.2P(A \cap B) = 0.2P(A∩B)=0.2, are A and B independent?

Correct Answer: A

Q16.A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is a spade or an ace?

Correct Answer: D

Q17.If two events AAA and BBB are independent, which of the following is true?

Correct Answer: B

Q18.A factory has two machines, A and B. Machine A produces 60% of the items, and 2% are defective. Machine B produces 40% of the items, and 1% are defective. What is the probability that a randomly selected defective item was produced by Machine B?

Correct Answer: C

Q19.A die is rolled. What is the probability of getting an even number?

Correct Answer: C

Q20.A die is rolled twice. What is the probability that the sum of the numbers is 8, given that the first roll is a 3?

Correct Answer: B

Q21.Two events AAA and BBB are such that P(A)=0.3P(A) = 0.3P(A)=0.3, P(B)=0.4P(B) = 0.4P(B)=0.4, and P(A∪B)=0.58P(A \cup B) = 0.58P(A∪B)=0.58. Are AAA and BBB independent?

Correct Answer: B

Q22.A die is thrown twice. Let EEE be the event that the sum of the numbers is 8, and FFF be the event that the first number is 3. What is the conditional probability P(E∣F)P(E|F)P(E∣F)?

Correct Answer: A

Q23.In a family with two children, what is the probability that both are girls given that the youngest is a girl?

Correct Answer: B

Q24.What is the probability of drawing a red ball from Bag II, given that Bag II contains 4 white and 5 red balls?

Correct Answer: B

Q25.If P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.3P(B) = 0.3P(B)=0.3, and P(A∩B)=0.1P(A \cap B) = 0.1P(A∩B)=0.1, what is the probability of AAA given BBB, P(A∣B)P(A|B)P(A∣B)?

Correct Answer: C

Q26.A factory has two machines, A and B. Machine A produces 60% of the items, and machine B produces 40%. If 2% of items from A and 1% from B are defective, what is the probability that a randomly selected defective item was produced by machine B?

Correct Answer: C

Q27.If P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.4P(B) = 0.4P(B)=0.4, and P(A∪B)=0.7P(A \cup B) = 0.7P(A∪B)=0.7, what is P(A∩B)P(A \cap B)P(A∩B)?

Correct Answer: B