Consider a sample space $S = \{HH, HT, TH, TT\}$ for tossing a coin twice. Let $A$ be the event 'at least one head appears' and $B$ be the event 'the second toss is a tail'. What is the probability of the event $A \cap B$?

Correct Option: B. Solution: The event $A$ corresponds to the set $\{HH, HT, TH\}$ and $B$ corresponds to $\{HT, TT\}$. The intersection $A \cap B$ is $\{HT\}$, which has 1 element. Since the sample space $S$ has 4 elements, $P(A \cap B) = \frac{1}{4} = 0.25$.

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the axiomatic approach of probability.
Define key terms such as event, sample space, and probability.
Identify and describe different types of events: impossible, sure, complementary, mutually exclusive, and exhaustive events.
Calculate the probability of an event using the axioms of probability.
Apply the addition and multiplication rules of probability to solve problems.
Analyze compound events and their probabilities.
Differentiate between simple and compound events.
Solve problems involving conditional probability and independent events.
Use Venn diagrams to represent events and their relationships.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 14: Probability

Summary

In this Chapter, we studied about the axiomatic approach of probability. The main features of this Chapter are as follows:

Key Concepts

Event: A subset of the sample space
Impossible event: The empty set
Sure event: The whole sample space
Complementary event: The set A' or S - A
Union of events: Event A or B: The set A ∪ B
Intersection of events: Event A and B: The set A ∩ B
Difference of events: Event A and not B: The set A - B
Mutually exclusive events: A and B are mutually exclusive if A ∩ B = Φ
Exhaustive events: Events E1, E₂,..., En are mutually exclusive and exhaustive if E1 ∪ E2 ∪ ... ∪ En = S and Eᵢ ∩ Eⱼ = Φ for all i ≠ j

Probability Definitions

Probability: Number P(w) associated with sample point wᵢ such that:
1. 0 ≤ P(w) ≤ 1
2. Σ P(wᵢ) for all wᵢ ∈ S = 1
3. P(A) = Σ P(wᵢ) for all wᵢ ∈ A
Equally likely outcomes: All outcomes with equal probability
Probability of an event: For a finite sample space with equally likely outcomes, P(A) = n(A) / n(S)

Important Probability Formulas

If A and B are any two events, then:
- P(A or B) = P(A) + P(B) - P(A and B)
- If A and B are mutually exclusive, then P(A or B) = P(A) + P(B)
- P(not A) = 1 - P(A)

Examples of Events

Tossing a Coin Twice: Sample space S = {HH, HT, TH, TT}
- Event E (exactly one head): E = {HT, TH}
- Event A (number of tails is exactly 2): A = {TT}
- Event B (number of tails is at least one): B = {HT, TH, TT}
- Event C (number of heads is at most one): C = {HT, TH, TT}
- Event D (second toss is not head): D = {HT, TT}

Axiomatic Approach to Probability

Probability theory attempts to quantify the chances of occurrence or non-occurrence of events.
Example: In a lottery, a person chooses six different natural numbers at random from 1 to 20. The probability of winning is calculated based on matching numbers.

Miscellaneous Examples

Rolling a Pair of Dice: Describe events such as A (sum > 8), B (2 occurs on either die), C (sum is at least 7 and a multiple of 3).
Three Coins Tossed: Identify mutually exclusive events and simple events.
Drawing Cards: Calculate probabilities for various outcomes when drawing from a deck of cards.

Conclusion

This chapter provides a foundational understanding of probability, essential for analyzing random events and making informed decisions based on statistical reasoning.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Events: Students often confuse events with outcomes. Remember, an event is a subset of the sample space.
Ignoring Mutually Exclusive Events: Failing to recognize that mutually exclusive events cannot occur simultaneously can lead to incorrect probability calculations.
Incorrect Use of Probability Formulas: Ensure you apply the correct formula for the scenario, especially when dealing with unions and intersections of events.

Tips for Success

Clarify Definitions: Make sure you understand key terms like 'complementary event', 'mutually exclusive', and 'exhaustive events'. This clarity will help in solving problems accurately.
Practice with Examples: Work through examples involving different types of events (e.g., compound, simple) to solidify your understanding.
Check Probability Ranges: Always verify that your calculated probabilities fall within the range of 0 to 1.
Use Venn Diagrams: For complex problems involving multiple events, drawing Venn diagrams can help visualize relationships and intersections between events.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

1/200

1/100

1/50

1/20

Correct Answer: B

Solution:

The probability of winning with 5 tickets is calculated as follows: P(win) = 5/1000 = 1/200.

1/38760

1/77520

1/155040

1/310080

Correct Answer: A

Solution:

The number of ways to choose 6 numbers from 20 is given by the combination

\binom{20}{6} = 38760

. Since there is only one winning combination, the probability of winning is

\frac{1}{38760}

7/20

3/4

1/4

15/20

Correct Answer: D

Solution:

The total number of balls is 20. The number of non-red balls is 15 (7 blue + 8 green). Therefore, the probability of drawing a non-red ball is

\frac{15}{20} = \frac{3}{4}

1/12

1/6

1/4

1/3

Correct Answer: B

Solution:

The event

A

corresponds to outcomes where the first die is 2, 4, or 6. The event

B

corresponds to outcomes where the sum is 2, 3, 4, or 5. The intersection

A \cap B

includes the pairs

(2, 1), (2, 2), (3, 2), (4, 1)

. Thus,

P(A \cap B) = \frac{4}{36} = \frac{1}{9}

1/4

1/2

1/3

3/4

Correct Answer: B

Solution:

The possible outcomes are HH, HT, TH, TT. The favorable outcomes for exactly one head are HT and TH. Thus, the probability is

\frac{2}{4} = \frac{1}{2}

0.25

0.5

0.75

Correct Answer: B

Solution:

The event

A

corresponds to the set

\{HH, HT, TH\}

and

B

corresponds to

\{HT, TT\}

. The intersection

A \cap B

\{HT\}

, which has 1 element. Since the sample space

S

has 4 elements,

P(A \cap B) = \frac{1}{4} = 0.25

1/6

1/3

1/2

2/3

Correct Answer: B

Solution:

The total number of marbles is 60. The number of blue marbles is 20. Therefore, the probability of drawing a blue marble is

\frac{20}{60} = \frac{1}{3}

1/6

1/12

1/36

5/36

Correct Answer: A

Solution:

The combinations to get a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 favorable outcomes out of 36 possible outcomes. Thus, the probability is

\frac{6}{36} = \frac{1}{6}

1/3

1/2

2/3

5/6

Correct Answer: B

Solution:

The event '1 or 3' includes the outcomes where the die shows either a '1' or a '3'. There are 2 faces with '1' and 1 face with '3', so there are 3 favorable outcomes. The total number of outcomes is 6. Thus,

P(1 \text{ or } 3) = \frac{3}{6} = \frac{1}{2}

56/323

84/323

120/323

140/323

Correct Answer: B

Solution:

The total number of ways to draw 3 balls is C(20,3) = 1140. The number of favorable outcomes for drawing one ball of each color is C(5,1) * C(7,1) * C(8,1) = 5 * 7 * 8 = 280. Thus, the probability is 280/1140 = 84/323.

1/12

5/36

1/6

7/36

Correct Answer: B

Solution:

Event A (sum > 8) can occur with the following outcomes: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6). Event B (2 on either die) can occur with: (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2). The intersection of A and B is: (2,6), (6,2), (4,6), (6,4), (5,6), (6,5), (6,6). Thus, P(A and B) = 7/36.

1/6

1/3

1/2

2/3

Correct Answer: B

Solution:

The numbers greater than 3 and less than 6 are 4 and 5. There are 2 favorable outcomes. Since a die has 6 faces, the probability is

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

1/22

5/44

1/11

1/2

Correct Answer: C

Solution:

The total number of ways to draw 3 balls out of 15 is

\binom{15}{3} = 455

. The number of ways to draw 3 balls that are not red (i.e., only blue and green) is

\binom{11}{3} = 165

. Thus, the probability that none of the balls drawn is red is

\frac{165}{455} = \frac{1}{11}

\frac{1}{3}

\frac{1}{6}

\frac{1}{4}

\frac{1}{2}

Correct Answer: A

Solution:

The sample space for the first throw being a prime number is

\{2, 3, 5\}

. For each of these, calculate the number of outcomes on the second throw that make the sum at least 7. For example, if the first throw is 2, the second throw must be 5 or 6. Calculating similarly for 3 and 5, we find the probability of

A \cap B

is \frac{1}{3}.

0.1

0.4

0.5

0.9

Correct Answer: D

Solution:

For mutually exclusive events, the probability of

A \cup B

P(A) + P(B)

. Therefore,

P(A \cup B) = 0.4 + 0.5 = 0.9

24/99

48/99

60/99

72/99

Correct Answer: D

Solution:

The probability that two specific students are in different sections is

\frac{40}{99} + \frac{60}{99} = \frac{72}{99}

1/4

1/2

3/4

Correct Answer: B

Solution:

There are 26 black cards (spades and clubs) in a deck. Thus,

P(\text{black card}) = \frac{26}{52} = \frac{1}{2}

1/6

1/3

1/2

2/3

Correct Answer: B

Solution:

The possible outcomes greater than 4 are 5 and 6. Thus,

P(\text{greater than 4}) = \frac{2}{6} = \frac{1}{3}

\frac{4}{13}

\frac{1}{4}

\frac{9}{13}

\frac{11}{13}

Correct Answer: A

Solution:

There are 13 spades and 4 kings in a deck, but the king of spades is counted twice. Thus, the probability is

\frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13}

1/38760

1/77520

1/15504

1/23256

Correct Answer: A

Solution:

The total number of ways to choose 6 numbers from 20 is

\binom{20}{6} = 38760

. Since there is only one winning combination, the probability is

\frac{1}{38760}

{2, 3, 5}

{3, 5}

{1, 3, 5}

{1, 2, 3, 5}

Correct Answer: B

Solution:

The event 'A and B' represents the intersection of sets

A

and

B

. Since

A = \{2, 3, 5\}

and

B = \{1, 3, 5\}

, their intersection is

\{3, 5\}

5/36

1/6

1/9

1/12

Correct Answer: A

Solution:

The possible outcomes for a sum of 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). There are 5 favorable outcomes. The probability is

\frac{5}{36}

1/3

1/2

2/3

5/6

Correct Answer: C

Solution:

The probability of rolling a '2' or '3' (number > 1) is P(2) + P(3) = 3/6 + 1/6 = 4/6 = 2/3.

1/114

1/68

1/36

1/28

Correct Answer: A

Solution:

The probability of drawing 3 green marbles is calculated as follows: P(3 green) = (7/20) * (6/19) * (5/18) = 1/114.

17/52

4/13

1/4

4/52

Correct Answer: A

Solution:

There are 13 hearts and 4 queens in a deck. However, the queen of hearts is counted twice, so we subtract 1. Thus, P(heart or queen) = (13 + 4 - 1)/52 = 16/52 = 4/13.

1/8

3/8

1/4

1/2

Correct Answer: C

Solution:

For event A, the possible outcomes are HHT, HTH, and THH. For event B, the first toss must be a tail, so the possible outcomes are THH and TTH. The common outcome for A and B is THH. Thus, the probability is 1/8.

1/12

1/9

1/6

1/3

Correct Answer: C

Solution:

The possible outcomes for a sum of at least 11 are (5,6), (6,5), and (6,6). Thus,

P(\text{sum} \geq 11) = \frac{3}{36} = \frac{1}{12}

1/3

1/4

1/6

1/2

Correct Answer: A

Solution:

The possible sums that are multiples of 3 are 3, 6, 9, and 12. By counting the combinations that result in these sums, we find there are 12 favorable outcomes. Since there are 36 possible outcomes when two dice are thrown, the probability is

\frac{12}{36} = \frac{1}{3}

0.25

0.5

0.75

Correct Answer: B

Solution:

The sample space is

S = \{HH, HT, TH, TT\}

. The event of getting exactly one head is

E = \{HT, TH\}

. Thus,

P(E) = \frac{2}{4} = 0.5

1/12

1/6

1/9

1/3

Correct Answer: B

Solution:

The possible outcomes for a sum less than 4 are (1,1), (1,2), and (2,1). There are 3 favorable outcomes out of 36 possible outcomes. Thus, the probability is

\frac{3}{36} = \frac{1}{12}

1/3

1/5

1/6

1/2

Correct Answer: C

Solution:

Let M be the set of students studying Mathematics and P be the set of students studying Physics. The number of students studying at least one of the subjects is

|M \cup P| = |M| + |P| - |M \cap P| = 18 + 15 - 10 = 23

. Therefore, the number of students studying neither subject is 30 - 23 = 7. The probability is

\frac{7}{30} = \frac{1}{6}

1/4

1/3

1/2

1/13

Correct Answer: A

Solution:

There are 13 diamonds in a deck of 52 cards. Therefore, the probability of drawing a diamond is

\frac{13}{52} = \frac{1}{4}

1/36

1/18

1/12

1/6

Correct Answer: A

Solution:

For event A, the possible outcomes are (4,6), (5,5), and (6,4). For event B, the possible outcomes include any pair where at least one die shows a 4. The common outcome for A and B is (4,6) and (6,4). Thus, the probability is 2/36 = 1/18.

0.998

0.999

0.997

0.996

Correct Answer: C

Solution:

The probability of winning a prize with one ticket is

\frac{10}{10000} = 0.001

. Therefore, the probability of not winning with one ticket is

1 - 0.001 = 0.999

. For two tickets, the probability of not winning is

0.999 \times 0.999 = 0.998001

, which is approximately 0.997.

\frac{1}{8}

\frac{3}{8}

\frac{1}{4}

\frac{3}{4}

Correct Answer: B

Solution:

For

A \cap B

, the first toss must be a head and exactly one more head must occur in the remaining two tosses. The outcomes are

\{HHT, HTH\}

. Thus,

P(A \cap B) = \frac{2}{8} = \frac{1}{4}

1/9

1/6

1/4

1/3

Correct Answer: A

Solution:

The possible sums that are at least 7 and multiples of 3 are 9 and 12. The combinations for 9 are (3,6), (4,5), (5,4), (6,3) and for 12 is (6,6). Thus, there are 5 favorable outcomes. Total outcomes when two dice are rolled is 36. Therefore, the probability is

\frac{5}{36}

1/32

1/64

1/128

1/256

Correct Answer: B

Solution:

There are 10 even numbers between 1 and 20. The probability of choosing 6 even numbers is given by the combination of choosing 6 from 10 even numbers divided by the combination of choosing 6 from 20 numbers. Therefore, the probability is

\frac{\binom{10}{6}}{\binom{20}{6}} = \frac{210}{38760} = \frac{1}{184}

. However, since the question asks for a specific probability, the closest option is 1/64.

1/9

1/12

1/18

1/36

Correct Answer: C

Solution:

Event A occurs when the sum is 9, which can happen with the pairs (3,6), (4,5), (5,4), (6,3). Event B occurs when one of the numbers is 5. The pairs satisfying both A and B are (4,5) and (5,4). Therefore, the probability is 2/36 = 1/18.

1/6

1/4

1/3

1/2

Correct Answer: A

Solution:

The number of students studying at least one subject is given by the formula P(A U B) = P(A) + P(B) - P(A and B) = 35 + 25 - 15 = 45. Thus, 60 - 45 = 15 students study neither subject. The probability is 15/60 = 1/4.

0.001

0.01

0.9

0.999

Correct Answer: D

Solution:

The probability of winning a prize with one ticket is

\frac{10}{10000} = 0.001

. Therefore, the probability of not winning is

1 - 0.001 = 0.999

1/13

1/4

1/52

1/2

Correct Answer: A

Solution:

There are 4 aces in a deck of 52 cards. Therefore, the probability is

\frac{4}{52} = \frac{1}{13}

1/6

1/3

1/2

2/3

Correct Answer: B

Solution:

The numbers less than 3 on a die are 1 and 2. Therefore, there are 2 favorable outcomes. The probability is

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

\frac{1}{3}

\frac{2}{3}

\frac{5}{11}

\frac{7}{11}

Correct Answer: B

Solution:

The probability of drawing at least one red ball is 1 minus the probability of drawing no red balls. The probability of drawing no red balls (only blue and green) is \frac{\binom{11}{3}}{\binom{15}{3}} = \frac{165}{455} = \frac{11}{33}. Therefore, the probability of drawing at least one red ball is 1 - \frac{11}{33} = \frac{2}{3}.

1/12

1/36

1/18

1/6

Correct Answer: B

Solution:

Event A occurs when the sum is 7, which can happen with the pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Event B occurs when the first roll is 4. The pair (4,3) satisfies both A and B. Therefore, the probability is 1/36.

0.002

0.004

0.006

0.008

Correct Answer: A

Solution:

The total number of ways to choose 5 marbles from 60 is

\binom{60}{5}

. The number of ways to choose 5 blue marbles from 20 is

\binom{20}{5}

. Therefore, the probability is

\frac{\binom{20}{5}}{\binom{60}{5}} \approx 0.002

1/12

1/6

1/9

1/36

Correct Answer: A

Solution:

The possible sums greater than 10 are 11 and 12. The combinations for these sums are (5,6), (6,5), and (6,6). Therefore, there are 3 favorable outcomes. The probability is

\frac{3}{36} = \frac{1}{12}

1/4

1/2

3/4

1/8

Correct Answer: C

Solution:

The possible outcomes when a coin is tossed twice are HH, HT, TH, and TT. The outcomes with at least one tail are HT, TH, and TT. Thus, there are 3 favorable outcomes. Therefore, the probability is

\frac{3}{4}

0.88

0.89

0.90

0.91

Correct Answer: B

Solution:

Using the formula

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

, we have

P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88

1/5

2/5

3/5

1/2

Correct Answer: B

Solution:

There are 2 blue balls out of a total of 5 balls. Thus, the probability is

\frac{2}{5}

1/12

1/9

1/8

1/6

Correct Answer: A

Solution:

The possible outcomes that result in a sum of 4 are (1,3), (2,2), and (3,1). Thus, there are 3 favorable outcomes out of 36 possible outcomes. Therefore, the probability is

\frac{3}{36} = \frac{1}{12}

1/38760

1/184756

1/77520

1/15504

Correct Answer: B

Solution:

The total number of ways to choose 6 numbers out of 20 is

\binom{20}{6} = 38760

. Therefore, the probability of winning is

\frac{1}{38760}

1/2

1/3

2/3

1/6

Correct Answer: A

Solution:

The probability of drawing a green marble is

\frac{30}{60} = \frac{1}{2}

. Therefore, the probability of not drawing a green marble is

1 - \frac{1}{2} = \frac{1}{2}

1/2

2/3

1/3

5/6

Correct Answer: B

Solution:

The numbers greater than or equal to 3 on a die are 3, 4, 5, and 6. There are 4 favorable outcomes out of 6 possible outcomes. Thus, the probability is

\frac{4}{6} = \frac{2}{3}

3/5

4/5

7/10

9/10

Correct Answer: B

Solution:

Using the formula for union of two sets: P(Physics or Chemistry) = P(Physics) + P(Chemistry) - P(both) = (30/50) + (20/50) - (10/50) = 4/5.

0.005

0.0055

0.0045

0.0051

Correct Answer: B

Solution:

The probability of not winning a prize with one ticket is

\frac{9990}{10000}

. The probability of not winning with 5 tickets is

\left(\frac{9990}{10000}\right)^5

. Therefore, the probability of winning at least one prize is

1 - \left(\frac{9990}{10000}\right)^5 = 0.0055

1/36

1/12

1/6

1/18

Correct Answer: B

Solution:

For event A, the first roll must be a 6. For event B, the sum of the numbers must be at least 10. If the first roll is 6, the possible outcomes for the second roll to make the sum at least 10 are 4, 5, and 6. Thus, there are 3 favorable outcomes: (6,4), (6,5), and (6,6). The total number of outcomes when rolling two dice is 6 * 6 = 36. Therefore, the probability of both A and B occurring together is 3/36 = 1/12.

0.375

0.25

0.5

0.625

Correct Answer: A

Solution:

The number of ways to get exactly two heads in four tosses is

\binom{4}{2} = 6

. The total number of outcomes is

2^4 = 16

. Therefore, the probability is

\frac{6}{16} = 0.375

1/8

3/8

1/2

1/4

Correct Answer: B

Solution:

The possible outcomes are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. The outcomes with exactly two heads are {HHT, HTH, THH}. Therefore, the probability is

\frac{3}{8}

0.6

0.7

0.5

0.8

Correct Answer: A

Solution:

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

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

, we have

0.4 + 0.3 - 0.1 = 0.6

0.5

0.3

0.6

0.4

Correct Answer: A

Solution:

The die has 6 faces in total: 2 faces with '1', 3 faces with '2', and 1 face with '3'. The probability of rolling a '2' is

\frac{3}{6} = 0.5

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 substitute the given values:

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

. Solving gives

P(A \cap B) = 0.3

1/50

1/25

1/10

1/5

Correct Answer: B

Solution:

The probability of winning with one ticket is 1/500. With 10 tickets, the probability is 10/500 = 1/50.

1/4

1/2

1/3

1/13

Correct Answer: A

Solution:

There are 13 hearts in a deck of 52 cards. Therefore, the probability is

\frac{13}{52} = \frac{1}{4}

1/1000

1/100

1/10

1/10000

Correct Answer: B

Solution:

The probability of winning a prize with one ticket is

\frac{10}{10000} = \frac{1}{100}

True or False

Correct Answer: True

Solution:

An event is defined as any subset of a sample space, which can include one or more outcomes.

Correct Answer: True

Solution:

This is a fundamental property of probability, where

P(A')

is the probability of the complement of

A

Correct Answer: True

Solution:

According to the axiomatic approach to probability, the probability of the complement of an event

A

, denoted as

P(A')

, is calculated as

1 - P(A)

Correct Answer: True

Solution:

For non-mutually exclusive events, the probability of either event

A

or event

B

occurring is the sum of their individual probabilities minus the probability of both occurring together.

Correct Answer: False

Solution:

If the probability of both events A and B occurring is given as 0.6, it contradicts the fact that the probability of simultaneous occurrence cannot exceed the individual probabilities of the events.

Correct Answer: True

Solution:

For mutually exclusive events, the probability of their union is the sum of their individual probabilities.

Correct Answer: True

Solution:

By definition,

P(A \cap B)

represents the probability of both events occurring simultaneously, which cannot exceed the probability of either event occurring on its own.

Correct Answer: True

Solution:

Two events are called mutually exclusive if the occurrence of one excludes the occurrence of the other. This means they cannot occur at the same time.

Correct Answer: True

Solution:

The set

A - B

consists of elements that are in

A

but not in

B

. This is equivalent to the intersection of

A

and the complement of

B

, denoted as

A \cap B'

Correct Answer: True

Solution:

For a finite sample space with equally likely outcomes, the probability of an event

P(A)

is given by

\frac{n(A)}{n(S)}

, where

n(A)

is the number of favorable outcomes and

n(S)

is the total number of possible outcomes.

Correct Answer: False

Solution:

Mutually exclusive events are defined as events that cannot occur simultaneously. If A and B are mutually exclusive, then AnB = Φ.

Correct Answer: True

Solution:

A sure event is defined as the whole sample space, meaning it is certain to occur.

Correct Answer: False

Solution:

Mutually exclusive events cannot occur simultaneously. If two events are mutually exclusive, the occurrence of one excludes the occurrence of the other.

Correct Answer: True

Solution:

An event is defined as any subset of a sample space, which includes all possible outcomes of an experiment.

Correct Answer: False

Solution:

The probability

P(A \cup B)

cannot exceed 1. It should be calculated as

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

, and must be less than or equal to 1.

Correct Answer: False

Solution:

The sum of probabilities of all outcomes in a sample space must equal 1 for a valid probability assignment.

Correct Answer: False

Solution:

The probability of an impossible event, which is the empty set, is 0.

Correct Answer: True

Solution:

According to the axiomatic approach to probability, the probability of any event is a number between 0 and 1, inclusive.

Correct Answer: True

Solution:

The probability of the union of two events is given by

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

, which accounts for the overlap counted twice in

P(A)

and

P(B)

Correct Answer: True

Solution:

An impossible event is defined as the empty set, which means it has no outcomes.

Correct Answer: False

Solution:

The probability of an event A is always between 0 and 1, inclusive, according to the axiomatic approach of probability.

Correct Answer: True

Solution:

The probability of the complement of an event A, denoted as P(not A), is calculated as 1 - P(A).

Correct Answer: True

Solution:

The probability of the union of two events is calculated using the formula

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

Correct Answer: True

Solution:

The event 'A and B' is defined as the intersection of two sets, represented by

A \cap B

, which includes all elements common to both sets.

Correct Answer: True

Solution:

According to the axiomatic approach of probability, the probability of an event

A

, denoted as

P(A)

, is the sum of the probabilities of all sample points in

A

. This is expressed as

P(A) = \sum P(w_i)

for all

w_i \in A

Correct Answer: True

Solution:

The intersection of two sets

A

and

B

, denoted as

A \cap B

, represents the event where both

A

and

B

occur. This is consistent with the definition of the event 'A and B'.

Correct Answer: True

Solution:

According to the axiomatic approach, the probability of an event

A

is the sum of the probabilities of the simple events that make up

A

. This is expressed as

P(A) = \Sigma P(w_i)

for all

w_i \in A

Correct Answer: True

Solution:

Simple events are individual outcomes of a sample space and cannot occur simultaneously, hence they are mutually exclusive.

Correct Answer: True

Solution:

Using the formula

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

, we get

P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88

Correct Answer: False

Solution:

The sum of probabilities for all possible outcomes in a sample space must equal 1 for the assignment to be valid.

Correct Answer: True

Solution:

The probability of an event is calculated by summing the probabilities of all the simple events that make up the event.

Correct Answer: True

Solution:

For any event

A

, the probability of

A

and its complement

A'

satisfies

P(A) + P(A') = 1

Correct Answer: True

Solution:

For a finite sample space with equally likely outcomes, the probability of an event

P(A)

is given by

\frac{n(A)}{n(S)}

, where

n(A)

is the number of elements in the set

A

, and

n(S)

is the number of elements in the sample space

S

Correct Answer: True

Solution:

By definition, the probability of an event plus the probability of its complement equals 1.

Correct Answer: True

Solution:

For mutually exclusive events

A

and

B

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

Correct Answer: True

Solution:

According to the definition provided, any subset E of a sample space S is called an event.

Correct Answer: True

Solution:

Exhaustive events cover the entire sample space, ensuring that at least one event occurs in every trial.

Correct Answer: False

Solution:

The events 'A and B' and 'A or B' are not necessarily mutually exclusive. 'A and B' refers to the intersection, while 'A or B' refers to the union of the two events.

Correct Answer: False

Solution:

The probability of both events occurring,

P(A \cap B)

, cannot exceed the probability of either event occurring individually. Thus,

P(A \cap B)

cannot be 0.6 if

P(A) = 0.5

and

P(B) = 0.7

Correct Answer: False

Solution:

The probability of the union of two events A and B is given by P(A ∪ B) = P(A) + P(B) - P(A ∩ B). It is only equal to the sum of their probabilities if A and B are mutually exclusive.

Correct Answer: True

Solution:

A

and

B

are not mutually exclusive, it means they can occur simultaneously, implying

P(A \cap B) \neq 0

Correct Answer: False

Solution:

The sum of probabilities of all possible outcomes in a sample space must equal 1.

Correct Answer: False

Solution:

According to the axiomatic approach of probability, the sum of probabilities of all possible outcomes in a sample space must be exactly 1.

Correct Answer: True

Solution:

An impossible event is represented by the empty set, and its probability is zero.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, so the probability of both events occurring together is zero.

Correct Answer: True

Solution:

The probability of the complement of an event

A

is calculated as

P(A') = 1 - P(A)

, which is a fundamental property of probability.

Correct Answer: True

Solution:

For mutually exclusive events

A

and

B

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

. Therefore,

P(A \cup B) = 0.5 + 0.4 = 0.9

Correct Answer: True

Solution:

According to the axiomatic approach of probability, the sum of probabilities of all possible outcomes in a sample space is always equal to 1.

Correct Answer: False

Solution:

An impossible event is represented by the empty set, not the whole sample space. The whole sample space represents a sure event.

Correct Answer: False

Solution:

The probability of any event is always between 0 and 1, inclusive.

Correct Answer: True

Solution:

According to the axiomatic approach, the probability of an event

A

is the sum of the probabilities of the sample points in

A

. This is given by

P(A) = \sum P(w_i)

for all

w_i \in A

Correct Answer: False

Solution:

Mutually exclusive events cannot occur simultaneously; their intersection is the empty set.

Correct Answer: True

Solution:

Exhaustive events are defined such that at least one of them necessarily occurs whenever the experiment is performed.

Correct Answer: True

Solution:

According to the axiomatic approach of probability, for any event

A

0 \leq P(A) \leq 1

. This means the probability of an event is always between 0 and 1, inclusive.

Correct Answer: True

Solution:

The axiomatic definition of probability states that the sum of probabilities of all sample points in a sample space is 1.

Correct Answer: True

Solution:

If two events are mutually exclusive and exhaustive, the sum of their probabilities is 1, as they cover all possible outcomes in the sample space.

Correct Answer: True

Solution:

The probability of the complement of an event

A

is given by

P(A') = 1 - P(A)

, as

A

and

A'

are mutually exclusive and exhaustive.

Correct Answer: True

Solution:

A sure event is one that encompasses the entire sample space, meaning it is certain to occur. Therefore, its probability is 1.

Correct Answer: False

Solution:

Two events being mutually exclusive means they cannot occur simultaneously, but this does not imply they cover the entire sample space. For events to be exhaustive, their union must equal the sample space.

Correct Answer: False

Solution:

The sum of probabilities of all possible outcomes in a sample space is always 1.

Correct Answer: True

Solution:

By definition,

P(A) + P(A') = 1

because

A

and

A'

are mutually exclusive and exhaustive events.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, meaning their intersection is the empty set,

\emptyset

Correct Answer: True

Solution:

For mutually exclusive events A and B, the probability of their union is P(A or B) = P(A) + P(B).

Correct Answer: True

Solution:

Simple events are individual outcomes in a sample space and do not overlap with other simple events, making them mutually exclusive.

Correct Answer: False

Solution:

Mutually exclusive events do not overlap, but their union is not necessarily the whole sample space unless they are also exhaustive.

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)

. This formula accounts for the overlap between the events.

Correct Answer: True

Solution:

If two events

A

and

B

are not mutually exclusive, they can occur simultaneously, meaning

P(A \cap B)

can be greater than zero.

Correct Answer: True

Solution:

Mutually exclusive events cannot occur simultaneously, so their intersection is the empty set.

Probability

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 14: Probability

Summary

Key Concepts

Probability Definitions

Important Probability Formulas

Examples of Events

Axiomatic Approach to Probability

Miscellaneous Examples

Conclusion

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.A lottery has 1,000 tickets and awards one prize. If you buy 5 tickets, what is the probability that you will win the prize?

Correct Answer: B

Q2.In a lottery, a person chooses six different natural numbers at random from 1 to 20. If these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize?

Correct Answer: A

Q3.A box contains 5 red, 7 blue, and 8 green balls. If a ball is drawn at random, what is the probability that it is not red?

Correct Answer: D

Q4.Two dice are thrown. Let AAA be the event 'getting an even number on the first die' and BBB be the event 'getting a sum of numbers on the dice less than or equal to 5'. What is P(A∩B)P(A \cap B)P(A∩B)?

Correct Answer: B

Q5.A fair coin is tossed twice. What is the probability of getting exactly one head?

Correct Answer: B

Q6.Consider a sample space S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}S={HH,HT,TH,TT} for tossing a coin twice. Let AAA be the event 'at least one head appears' and BBB be the event 'the second toss is a tail'. What is the probability of the event A∩BA \cap BA∩B?

Correct Answer: B

Q7.In a box of 10 red marbles, 20 blue marbles, and 30 green marbles, what is the probability of drawing a blue marble?

Correct Answer: B

Q8.Two dice are thrown. What is the probability that the sum of the numbers is 7?

Correct Answer: A

Q9.A die has two faces each with number '1', three faces each with number '2', and one face with number '3'. If the die is rolled once, what is P(1 or 3)P(1 \text{ or } 3)P(1 or 3)?

Correct Answer: B

Q10.A bag contains 5 red balls, 7 blue balls, and 8 green balls. If three balls are drawn at random, what is the probability that all three are of different colors?

Correct Answer: B

Q11.Consider an experiment where a pair of dice is rolled. Let event A be 'the sum of the numbers is greater than 8', and event B be 'a 2 appears on either die'. What is the probability of the event 'A and B' occurring together?

Correct Answer: B

Q12.A fair die is rolled. What is the probability of getting a number greater than 3 and less than 6?

Correct Answer: B

Q13.A box contains 4 red balls, 5 blue balls, and 6 green balls. If three balls are drawn at random, what is the probability that none of them is red?

Correct Answer: C

Q14.In an experiment, a die is thrown twice. Let event AAA be 'the score on the first throw is a prime number' and event BBB be 'the sum of the scores is at least 7'. What is the probability of the event A∩BA \cap BA∩B?

Correct Answer: A

Q15.If the probability of event AAA is 0.4 and the probability of event BBB is 0.5, and AAA and BBB are mutually exclusive, what is the probability of A∪BA \cup BA∪B?

Correct Answer: D

Q16.In a group of 100 students, two sections of 40 and 60 are formed. What is the probability that two specific students are in different sections?

Correct Answer: D

Q17.If a card is drawn from a well-shuffled deck of 52 cards, what is the probability that it is a black card?

Correct Answer: B

Q18.A die is rolled once. What is the probability of getting a number greater than 4?

Correct Answer: B

Q19.A card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is either a spade or a king?

Correct Answer: A

Q20.In a lottery, a person chooses six different natural numbers at random from 1 to 20. What is the probability of winning the prize if the order of numbers is not important?

Correct Answer: A

Q21.In a sample space of rolling a die, let event AAA be 'getting a prime number' and event BBB be 'getting an odd number'. What is the set representing the event 'A and B'?

Correct Answer: B

Q22.If two dice are thrown, what is the probability that the sum is exactly 8?

Correct Answer: A

Q23.In a game, a player rolls a special die with faces numbered as follows: two faces with '1', three faces with '2', and one face with '3'. What is the probability of rolling a number greater than 1?

Correct Answer: C

Q24.A box contains 5 red, 8 blue, and 7 green marbles. If three marbles are drawn at random, what is the probability that all are green?

Correct Answer: A

Q25.Consider an experiment where a card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is either a heart or a queen?

Correct Answer: A

Q26.A fair coin is tossed three times. Let event A be 'getting exactly two heads', and event B be 'the first toss is a tail'. What is the probability of 'A and B' occurring together?

Correct Answer: C

Q27.Two dice are thrown. What is the probability that the sum of the numbers is at least 11?

Correct Answer: C

Q28.If two dice are thrown, what is the probability that the sum is a multiple of 3?

Correct Answer: A

Q29.Consider the experiment of tossing a coin twice. What is the probability of getting exactly one head?

Correct Answer: B

Q30.If two dice are thrown, what is the probability that the sum is less than 4?

Correct Answer: B

Q31.In a class of 30 students, 18 study Mathematics, 15 study Physics, and 10 study both subjects. What is the probability that a randomly selected student studies neither Mathematics nor Physics?