Chapter 14 Probability

NCERT Solutions Class 11 Maths Chapter 14 – Probability

NCERT Solutions for Class 11 Maths Chapter 14 (Probability) deals with the mathematical study of uncertainty. Chapter 10 was purely frequency based. In this chapter we introduce the Axiomatic Approach to Probability. It is built on the concepts of Sets and Permutations & Combinations and gives us a formal structure to predict the probability of various outcomes of random experiments.

ALLEN’s NCERT Solutions for Class 11 Maths Chapter 14 is designed by expert faculty to help students to shift from intuitive guesswork to logical calculation. The exact definition of Sample Spaces and the classification of Events are the building blocks of the more advanced "Conditional Probability" and "Bayes' Theorem" in Class 12, on which the solutions are based.

Probability is an important topic for JEE Main, JEE Advanced and NDA exams. These solutions deals with problems on mutually exclusive and exhaustive events which are important topics in mathematics and data science sections in high level entrance exams.

1.0Class 11 Maths Chapter 14: Key Concepts

This chapter focuses on the formalization of random experiments and the quantification of their outcomes. Key lessons include:

Random Experiments: Experiments where the result cannot be predicted with certainty, even though all possible outcomes are known.
Sample Space (S): The set of all possible outcomes of a random experiment.
Events: A subset of the sample space.
- Impossible and Sure Events: $ϕ$ and S.
- Simple Event: An event containing only one sample point.
- Compound Event: An event containing more than one sample point.
Algebra of Events: Using set operations to define new events:
- Complementary Event (A'): "Not A".
- Union of Events ( $A \cup B$ ): "A or B".
- Intersection of Events ( $A \cap B$ ): "A and B".
Mutually Exclusive Events: Events that cannot occur simultaneously ( $A \cap B = ϕ$ ).
Exhaustive Events: Events whose union covers the entire sample space.
Axiomatic Approach to Probability: Assigning probabilities based on specific rules ( $0 \leq P (A) \leq 1 an d P (S) = 1$ ).
Probability of 'A or B': $P (A \cup B) = P (A) + P (B) - P (A \cap B) .$

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 8

EXERCISE - 14.1

A die is rolled. Let E be the event "die shows 4" and F be the event "die shows even number".
Are E and F mutually exclusive?
Sol. When a die is rolled, the sample space is given by $S = {1, 2, 3, 4, 5, 6}$
Accordingly, $E = {4}$ and $F = {2, 4, 6}$
It is observed that $E \cap F = {4} \neq = ϕ$
Therefore, E and F are not mutually exclusive events.
A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3

Also find $A \cup B, A \cap B, B \cup C, E \cap F$ , $D \cap E, A - C, D - E, E \cap F^{'}, F^{'}$

Sol. When a die is thrown, the sample space is given by $S = {1, 2, 3, 4, 5, 6}$ .
Accordingly:
(i) $A = {1, 2, 3, 4, 5, 6}$
(ii) $B = ϕ$
(iii) $C = {3, 6}$
(iv) $D = {1, 2, 3}$
(v) $E = {6}$
(vi) $F = {3, 4, 5, 6}$

$A \cup B = {1, 2, 3, 4, 5, 6}, A \cap B = ϕ$
$A \cup C = {3, 6}, E \cap F = {6} D \cap E = ϕ$ , $A - C = {1, 2, 4, 5}, D - E = {1, 2, 3}$ , $F^{'} = {1, 2}, E \cap F^{'} = ϕ$

An experiment involves rolling a pair of dice and recording the numbers that come up.
Describe the following events:
A : the sum is greater than 8,
B : 2 occurs on either die
C : The sum is at least 7 and a multiple of 3 .
Which pairs of these events are mutually exclusive?
Sol. When a pair of dice is rolled, the sample space is given by

$S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6} = ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫$

Accordingly,

$A = B = C = {(3, 6), (4, 5), (4, 6), (5, 4) (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)} {(2, 1), (2, 2), (2, 3), (2, 4) (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} {(3, 6), (4, 5), (5, 4), (6, 3) (6, 6)}$

It is observed that

$A \cap B = ϕ B \cap C = ϕ C \cap A = {(3, 6), (4, 5), (5, 4), (6, 3) (6, 6)} \neq = ϕ$

Hence, events A and B and events B and C are mutually exclusive.

Three coins are tossed once. Let A denote the event "three heads show", B denote the event "two heads and one tail show". C denote the event "three tails show" and D denote the event "a head shows on the first coin". Which events are
(i) mutually exclusive?
(ii) simple?
(iii) compound?
Sol. When three coins are tossed, the sample space is given by
$S = {HHH, HHT, HTH, HTT, THH, THT$ , TTH, TTT}
Accordingly,
$A = {HHH}$
B = {HHT, HTH, THH}
$C = {TTT}$
$D = {HHH, HHT, HTH, HTT}$

We now observe that

$A \cap B = ϕ, A \cap C = ϕ, A \cap D = {HHH} \neq = ϕ B \cap C = ϕ, B \cap D = {HHT, {HTH} \neq = ϕ C \cap D = ϕ$

(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
Sol. When three coins are tossed, the sample space is given by
$S = {HHH, HHT, HTH, HTT, THH, THT$ , TTH, TTT}
(i) Two events that are mutually exclusive can be
A: getting no heads and B: getting no tails
This is because sets
$A = {TTT}$ and $B = {HHH}$ are disjoint.
(ii) Three events that are mutually exclusive and exhaustive can be
A : getting no heads
B : getting exactly one head
C : getting at least two heads
i.e., $A = {TTT}$
B = {HTT, THT, TTH}
$C = {HHH, HHT, HTH, THH}$
This is because $A \cap B = B \cap C = C \cap A = ϕ$ and $A \cup B \cup C = S$ (iii) Two events that are not mutually exclusive can be
A: getting three heads
B: getting at least 2 heads
i.e., $A = {HHH}$
B = {HHH, HHT, HTH, THH}
This is because $A \cap B = {HHH} \neq = ϕ$
(iv) Two events which are mutually exclusive but not exhaustive can be
A: getting exactly one head
B: getting exactly one tail That is

$A = {HTT, THT, TTH} B = {HHT, HTH, THH}$

It is because, $A \cap B = ϕ$ , but $A \cup B \neq = S$
(v) Three events that are mutually exclusive but not exhaustive can be
A : getting exactly three heads
B : getting one head and two tails
C : getting one tail and two heads
i.e., $A = {HHH}$
B = {HTT, THT, TTH}
C = {HHT, HTH, THH}
This is because $A \cap B = B \cap C = C \cap A = ϕ$ , but $A \cup B \cup C \neq = S$

Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\leq 5$
Describe the events
(i) A'
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C
(vii) B and C
(viii) $A^{'} \cap B^{'} \cap C^{'}$

Sol. When two dice are thrown, the sample space is given by

$S = ⎩ ⎨ ⎧ (x, y) : x, y = 1, 2, 3, 4, 5, 6} (2, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Accordingly, A = ⎩ ⎨ ⎧ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫ B = ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ⎭ ⎬ ⎫ C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}$

(i) $A^{'} = ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ⎭ ⎬ ⎫ = B$
(ii) Not $B = B^{'} ⎩ ⎨ ⎧ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫ = A$
(iii) A or $B = A \cup B$

$= ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫ = S$

(iv) A and $B = A \cap B = ϕ$
(v) A but not $C = A - C$

$= ⎩ ⎨ ⎧ (2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), ⎭ ⎬ ⎫$

(vi) B or $C = B \cup C$

$= ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ⎭ ⎬ ⎫$

(vii) B and $C = B \cap C$

${(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}$

(viii) $C^{'} = ⎩ ⎨ ⎧ (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫$
$∴ A \cap B^{'} \cap C^{'} = A \cap A \cap C^{'} = A \cap C^{'}$

$⎩ ⎨ ⎧ (2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫$

Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\leq 5$
State true or false: (give reason for your answer)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) $A = B^{'}$
(iv) A and C are mutually exclusive are mutually exclusive
(v) A and B' are mutually exclusive.
(vi) A', B' and Care mutually exclusive and exhaustive.

Sol. $A = ⎩ ⎨ ⎧ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫$

$B C = ⎩ ⎨ ⎧ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ⎭ ⎬ ⎫ = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}$

(i) It is observed that $A \cap B = ϕ$
$∴ A$ and B are mutually exclusive.
Thus, the given statement is true.
(ii) It is observed that $A \cap B = ϕ$ and $A \cup B = S$
$∴ A$ and B are mutually exclusive and exhaustive.
Thus, the given statement is true.
(iii) It is observed that

$B^{'} = ⎩ ⎨ ⎧ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) ⎭ ⎬ ⎫ = A$

Thus, the given statement is true.
(iv) It is observed that

$A \cap C = {(2, 1), (2, 2), (2, 3), (4, 1)} \neq = ϕ$

$∴ A$ and C are not mutually exclusive.
Thus, the given statement is false.
(v) $A \cap B^{'} = A \cap A = A$
$∵ A \cap B^{'} \neq = ϕ$
$∴ A$ and B ' are not mutually exclusive.
Thus, the given statement is false.
(vi) It can be bseved that: $A^{'} \cup B^{'} \cup C = S$

However,

$B^{'} \cap C = {(2, 1, (2, 2), (2, 3), (4, 1)} \neq = ϕ$

Therefore, events $A^{'}, B^{'}$ and Care not mutually exclusive and exhaustive.
Thus, the given statement is false.

EXERCISE - 14.2

Which of the following cannot be valid assignment of probabilities for outcomes of sample space

$S = {ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}, ω_{6}, ω_{7}}$

Assig.	$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
(a)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(b)	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$
(c)	0.1	0.2	0.3	0.4	0.5	0.6	0.7
(d)	-0.1	0.2	0.3	0.4	-0.2	0.1	0.3
(e)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Sol. (a)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
0.1	0.01	0.05	0.03	0.01	0.2	0.6

Here, each of the numbers $p (ω_{i})$ is positive and less than 1. Sum of probabilities

$= p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7}) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1$

Thus, the assignment is valid. (b)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$

Here, each of the numbers $p (ω_{i})$ is positive and less than 1 .

$= p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7}) = \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = 7 \times \frac{1}{7} = 1$

Thus, the assignment is valid. (c)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
0.1	0.2	0.3	0.4	0.5	0.6	0.7

Here, each of the numbers $p (ω_{i})$ is positive and less than 1.

Sum of probabilities

$= p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7}) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 \neq = 1$

Thus, the assignment is not valid.

(d)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
-0.1	0.2	0.3	0.4	-0.2	0.1	0.3

Here, $p (ω_{1})$ and $p (ω_{5})$ are negative. Hence, the assignment is not valid. (e)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Here, $p (ω_{7}) = \frac{15}{14} > 1$ Hence, the assignment is not valid.

A coin is tossed twice, what is the probability that at least one tail occurs? Sol. When a coin is tossed twice, the sample space is given by

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

Let A be the event of the occurrence of at least one tail. Accordingly, $A = {HT, TH, TT}$ $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

$= \frac{n ( A )}{n ( S )} = \frac{3}{4}$

A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear. Sol. The sample space of the given experiment is given by

$S = {1, 2, 3, 4, 5, 6}$

(i) Let A be the event of the occurrence of a prime number. Accordingly, $A = {2, 3, 5}$ $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

$= \frac{n ( A )}{n ( S )} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be the event of the occurrence of a number greater than or equal to 3 . Accordingly, $B = {3, 4, 5, 6}$ $∴ P (B) = \frac{Number of outcomes favourable to B}{Total number of possible outcomes}$

$= \frac{n ( B )}{n ( S )} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, $C = {1}$ $∴ P (C) = \frac{Number of outcomes favourable to C}{Total number of possible outcomes}$

$= \frac{n ( C )}{n ( S )} = \frac{1}{6}$

(iv) Let D be the event of the occurrence of a number greater than 6. Accordingly, $D = Φ$ $∴ P (D) = \frac{Number of outcomes favourable to D}{Total number of possible outcomes}$

$= \frac{n ( D )}{n ( S )} = \frac{0}{6} = 0$

(v) Let E be the event of the occurrence of a number less than 6 . Accordingly, $E = {1, 2, 3, 4, 5}$ $∴ P (E) = \frac{Number of outcomes favourable to E}{Total number of possible outcomes}$

$= \frac{n ( E )}{n ( S )} = \frac{5}{6}$

A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card.

Sol.

(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements. Therefore, there are 52 points in the sample space. (b) Let A be the event in which the card drawn is an ace of spades. Accordingly, $n (A) = 1$ $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

$= \frac{n ( A )}{n ( S )} = \frac{1}{52}$

(c) (i) Let E be the event in which the card drawn is an ace. Since there are 4 aces in a pack of 52 cards, $n (E) = 4$ $∴ P (E) = \frac{Number of outcomes favourable to E}{Total number of possible outcomes}$

$= \frac{n ( E )}{n ( S )} = \frac{4}{52} = \frac{1}{13}$

(ii) Let F be the event in which the card drawn is black. Since there are 26 black cards in a pack of 52 cards, $n (F) = 26$ $∴ P (F) = \frac{Number of outcomes favourable to F}{Total number of possible outcomes}$

$= \frac{n ( F )}{n ( S )} = \frac{26}{52} = \frac{1}{2}$

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Sol. Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by $S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$ , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Accordingly, $n (S) = 12$

(i) Let A be the event in which the sum of numbers that turn up is 3. Accordingly, $A = {(1, 2)}$ $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

$= \frac{n ( A )}{n ( S )} = \frac{1}{12}$

(ii) Let B be the event in which the sum of numbers that turn up is 12. Accordingly, $B = {(6, 6)}$ $∴ P (B) = \frac{Number of outcomes favourable to B}{Total number of possible outcomes}$

$= \frac{n ( B )}{n ( S )} = \frac{1}{12}$

There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? Sol. There are four men and six women on the city council. As one council member is to be selected for a committee at random, the sample space contains $10 (4 + 6)$ elements. Let A be the event in which the selected council member is a woman. Accordingly, $n (A) = 6$ $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

$= \frac{n ( A )}{n ( S )} = \frac{6}{10} = \frac{3}{5}$

A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Sol. Since the coin is tossed four times, there can be a maximum of 4 heads or tails. When 4 heads turns up, $Re 1 + Re 1 + Re 1 = Rs 4$ is the gain. When 3 heads and 1 tail turn up, $Re 1 + Re 1 + Re 1 - Rs 1.50 = Rs 3 - Rs 1.50$ $= Rs 1.50$ is the gain. When 2 heads and 2 tails turns up, $Re 1 + Re 1 - Rs 1.50 - Rs 1.50 = - Re 1$ , i.e., Re 1 is the loss. When 1 head and 3 tails turn up, Re 1 - Rs 1.50 - Rs 1.50 - Rs $1.50 = -$ Rs 3.50, i.e., Rs 3.50 is the loss. When 4 tails turn up, Rs 1.50 - Rs 1.50 - Rs 1.50 - Rs 1.50 $= -$ Rs 6.00 , i.e., Rs 6.00 is the loss. There are $2^{4} = 16$ elements in the sample space S, which is given by: $S = {HHHH, HHHT, HHTH, HTHH, THHH$ , HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT} $∴ n (S) = 16$

The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.

$∴$ Probability (of winning Rs 4.00) $= 1/16$ The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs. $∴$ Probability (of winning Rs 1.50 $) = 4/16 = 1/4$ The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs. $∴$ Probability (of losing Re 1.00) $= 6/16 = 3/8$ The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs. Probability (of losing Rs 3.50) $= 4/16 = 1/4$ The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs. Probability (of losing Rs 6.00) = 1/16

Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails. Sol. When three coins are tossed once, the sample space is given by $S = {HHH, HHT, HTH, THH, HTT, THT$ , TTH, TTT} Accordingly, $n (S) = 8$ It is known that the probability of an event A is given by $∴ P (A) = \frac{Number of outcomes favourable to A}{Total number of possible outcomes}$

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

(i) Let B be the event of the occurrence of 3 heads. Accordingly, $B = {HHH}$ $∴ P (B) = \frac{n ( B )}{n ( S )} = \frac{1}{8}$ (ii) Let C be the event of the occurrence of 2 heads. Accordingly, $C = {HHT, HTH, THH}$ $∴ P (C) = \frac{n ( C )}{n ( S )} = \frac{3}{8}$ (iii) Let D be the event of the occurrence of at least 2 heads. Accordingly, $D = {HHH, HHT, HTH, THH}$ $∴ P (D) = \frac{n ( D )}{n ( S )} = \frac{4}{8} = \frac{1}{2}$ (iv) Let E be the event of the occurrence of at most 2 heads. Accordingly, $E = {HHT, HTH, THH, HTT, THT, TTH, TTT}$ $∴ P (E) = \frac{n ( E )}{n ( S )} = \frac{7}{8}$ (v) Let F be the event of the occurrence of no head. Accordingly, $F = {TTT}$ $∴ P (F) = \frac{n ( F )}{n ( S )} = \frac{1}{8}$

(vi) Let G be the event of the occurrence of 3 tails. Accordingly, $G = {TTT}$ $∴ P (G) = \frac{n ( G )}{n ( S )} = \frac{1}{8}$ (vii) Let H be the event of the occurrence of exactly 2 tails. Accordingly, $H = {HTT, THT, TTH}$ $∴ P (H) = \frac{n ( H )}{n ( S )} = \frac{3}{8}$ (viii) Let I be the event of the occurrence of no tail. Accordingly, $I = {HHH}$ $∴ P (I) = \frac{n ( I )}{n ( S )} = \frac{1}{8}$ (ix) Let J be the event of the occurrence of at most 2 tails. Accordingly, $J = {HHH, HHT, HTH, THH$ , HTT, THT, TTH} $∴ P (J) = \frac{n ( J )}{n ( S )} = \frac{7}{8}$

If 2/11 is the probability of an event, what is the probability of the event 'not A'.

Sol. It is given that $P (A) = 2/11$ . Accordingly,

$P (not A) = 1 - P (A) = 1 - 2/11 = 9/11$

A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) a vowel (ii) an consonant Sol. There are 13 letters in the word ASSASSINATION. ∴ Hence, $n (S) = 13$ (i) There are 6 vowels in the given word. ∴ Probability $($ vowel $) = 6/13$ (ii) There are 7 consonants in the given word. $∴$ Probability $($ consonant $) = 7/13$
In a lottery, person choses six different natural numbers at random from 1 to 20, and 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 in the game? [Hint: order of the numbers is not important.] Sol. Total number of ways in which one can choose six different numbers from 1 to 20

$=^{20} C_{6} = \frac{20 !}{6 ! ( 20 - 6 )!} = \frac{20 !}{6 ! 14 !} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} = 38760$

Hence, there are 38760 combinations of 6 numbers. Out of these combinations, one combination is already fixed by the lottery committee. ∴ Required probability of winning the prize in the game $= \frac{1}{38760}$

12. Check whether the following probabilities P(A) and P(B) are consistently defined (i) $P (A) = 0.5, P (B) = 0.7, P (A \cap B) = 0.6$ (ii) $P (A) = 0.5, P (B) = 0.4, P (A \cup B) = 0.8$

Sol.

(i) $P (A) = 0.5, P (B) = 0.7, P (A \cap B) = 0.6$ It is known that if E and F are two events such that $E \subset F$ , then $P (E) \leq P (F)$ . However, here, $P (A \cap B) > P (A)$ . Hence, $P (A)$ and $P (B)$ are not consistently defined. (ii) $P (A) = 0.5, P (B) = 0.4, P (A \cup B) = 0.8$ It is known that if E and F are two events such that $E \subset F$ , then $P (E) \leq P (F)$ . Here, it is seen that $P (A \cup B) > P (A)$ and $P (A \cup B) > P (B)$ . Hence, $P (A)$ and $P (B)$ are consistently defined.

Fill in the blanks in following table:

	P(A)	P(B)	$P (A \cap B)$	$P (A \cup B)$
(i)	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	.....
(ii)	0.35	.....	0.25	0.6
(iii)	0.5	0.35	.....	0.7

Sol.

(i) Here, $P (A) = \frac{1}{3}, P (B) = \frac{1}{5}, P (A \cap B) = \frac{1}{15}$ We know that

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

$∴ P (A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} = \frac{5 + 3 - 1}{15} = \frac{7}{15}$ (ii) Here,

$P (A) = 0.35, P (A \cap B) = 0.25, P (A \cup B) = 0.6$

We know that

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

$∴ 0.6 = 0.35 + P (B) - 0.25$ $\Rightarrow P (B) = 0.6 - 0.35 + 0.25$ $\Rightarrow P (B) = 0.5$ (iii) Here, $P (A) = 0.5, P (B) = 0.35, P (A \cup B) = 0.7$ We know that

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

$∴ 0.7 = 0.5 + 0.35 - P (A \cap B)$ $\Rightarrow P (A \cap B) = 0.5 + 0.35 - 0.7$ $\Rightarrow P (A \cap B) = 0.15$

Given $P (A) = 3/5$ and $P (B) = 1/5$ . Find P(A or B), if A and B are mutually exclusive events. Sol. Here, $P (A) = 3/5, P (B) = 1/5$ For mutually exclusive events A and B,

$P (A or B) = P (A) + P (B)$

$∴ P (A$ or B $) = 3/5 + 1/5 = 4/5$

If E and F are events such that $P (E) = 1/4$ , $P (F) = 1/2$ and $P (E$ and F $) = 1/8$ , find: (i) P(E or F), (ii) P(not E and not F). Sol. Here, $P (E) = 1/4, P (F) = 1/2$ and $P (E$ and F $) = 1/8$ (i) We know that

$P (E or F) = P (E) + P (F) - P (E and F)$

$∴ P (E$ or F $) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2 + 4 - 1}{8} = \frac{5}{8}$ (ii) From (i), $P (E$ or F $) = P (E \cup F) = \frac{5}{8}$ We have $(E \cup F)^{'} = (E^{'} \cap F^{'})$ [By De Morgan's law] $∴ P (E^{'} \cap F^{'}) = P (E \cup F)^{'}$ Now, $P (E \cup F)^{'} = 1 - P (E \cup F) = 1 - \frac{5}{8} = \frac{3}{8}$ $∴ P (E^{'} \cap F^{'}) = \frac{3}{8}$ Thus, $P ($ not E and not F $) = \frac{3}{8}$

Events E and F are such that P(not E or not F) $= 0.25$ , State whether E and F are mutually exclusive. Sol. It is given that $P ($ not E or not F $) = 0.25$

$i.e., P (E^{'} \cap F^{'}) = 0.25$

$\Rightarrow P (E \cap F)^{'} = 0.25 [E^{'} \cup F^{'} = (E \cap F)^{'}]$ Now,

$P (E \cap F) = 1 - P (E \cap F)^{'}$

$\Rightarrow P (E \cap F) = 1 - 0.25$ $\Rightarrow P (E \cap F) = 0.75 \neq = 0$ $\Rightarrow E \cap F \neq = ϕ$

Thus, E and F are not mutually exclusive.

A and B are events such that $P (A) = 0.42$ , $P (B) = 0.48$ and $P (A$ and B $) = 0.16$ . Determine (i) P(not A), (ii) P (not B) and (iii) P(A or B). Sol. It is given that $P (A) = 0.42, P (B) = 0.48$ , $P (A$ and B $) = 0.16$ (i) $P ($ not A $) = 1 - P (A) = 1 - 0.42 = 0.58$ (ii) $P (not B) = 1 - P (B) = 1 - 0.48 = 0.52$ (iii) We know that

$P (A or B) = P (A) + P (B) - P (A and B)$

$∴ P (A$ or B $) = 0.42 + 0.48 - 0.16 = 0.74$

In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Sol. Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology. Now,

$P (A) = 40% = 40/100 = 2/5 P (B) = 30% = 30/100 = 3/10 P (A and B) = 10% = 10/100 = 1/10$

We know that

$∴ P (A or B) = P (A) + P (B) - P (A and B) P (A or B) = \frac{2}{5} + \frac{3}{8} - \frac{1}{10} = \frac{6}{10} = 0.6$

Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6.

In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both? Sol. Let A and B be the events of passing first and second examinations respectively. Accordingly,

$P (A) = 0.8, P (B) = 0.7 and P (A or B) = 0.95$

We know that

$P (A or B) = P (A) + P (B) - P (A and B)$

$∴ 0.95 = 0.8 + 0.7 - P (A$ and B $)$ $\Rightarrow P (A$ and B $) = 0.8 + 0.7 - 0.95 = 0.55$

Thus, the probability of passing both the examinations is 0.55.

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination? Sol. Let A and B be the events of passing English and Hindi examinations respectively. Accordingly, $P (A$ and B $) = 0.5, P ($ not A and not B $) = 0.1$ , i.e., $P (A^{'} \cap B^{'}) = 0.1, P (A) = 0.75$ Now, $P (A \cup B)^{'} = (A^{'} \cap B^{'})$ [De Morgan's law] $∴ P (A \cup B)^{'} = P (A^{'} \cap B^{'}) = 0.1$ $P (A \cup B) = 1 - P (A \cup B)^{'} = 1 - 0.1 = 0.9$ We know that $P (A$ or B $) = P (A) + P (B) - P (A$ and B $)$ $∴ 0.9 = 0.75 + P (B) - 0.5$ $\Rightarrow P (B) = 0.9 - 0.75 + 0.5$ $\Rightarrow P (B) = 0.65$

Thus, the probability of passing the Hindi examination is 0.65 .

In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC. Sol. Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS. Total number of students $= 60$ Number of students who have opted for $NCC = 30$ $∴ P (A) = 30/60 = 1/2$ Number of students who have opted for NSS = 32 $∴ P (B) = \frac{32}{60} = \frac{8}{15}$

Number of students who have opted for both NCC and NSS = 24

$∴ P (A$ and B $) = \frac{24}{60} = \frac{2}{5}$

(i) We know that

$P (A or B) = P (A) + P (B) - P (A and B)$

∴

$P (A or B) = \frac{1}{2} + \frac{8}{15} - \frac{2}{5} = \frac{15 + 16 - 12}{30} = \frac{19}{30}$

Thus, the probability that the selected student has opted for NCC or NSS is 19/30. (ii) P (not A and not B )

$= P (A^{'} and B^{'}) = P (A^{'} \cap B^{'}) = P (A \cup B)^{'} [(A^{'} \cap B^{'}) = (A \cup B)^{'} (by De Morgan’s law)] = 1 - P (A \cup B) = 1 - P (A or B) = 1 - \frac{19}{30} = \frac{11}{30}$

Thus, the probability that the selected students has neither opted for NCC nor NSS is 11/30. (iii) The given information can be represented by a Venn diagram as

It is clear that Number of students who have opted for NSS but not NCC

$= n (B - A) = n (B) - n (A \cap B) = 32 - 24 = 8$

Thus, the probability that the selected student has opted for NSS but not for NCC $= \frac{8}{60} = \frac{2}{15}$

MISCELLANEOUS EXERCISE

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? Sol. Total number of marbles $= 10 + 20 + 30 = 60$ Number of ways of drawing 5 marbles from 60 marbles $=^{60} C_{5}$ (i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles. 5 blue marbles can be drawn from 20 blue marbles in $^{20} C_{5}$ ways. ∴ Probability that all marbles will be blue $= \frac{^{20} C _{5}}{^{60} C _{5}}$ (ii) Number of ways in which the drawn marble is not green $=^{(20 + 10)} C_{5}$ $∴$ Probability that no marble is green $= \frac{^{30} C _{5}}{^{60} C _{5}}$ Probability that at least one marble is green $= 1 - \frac{^{30} C _{5}}{^{60} C _{5}}$
4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade? Sol. Number of ways of drawing 4 cards from 52 cards $=^{52} C_{4}$ In a deck of 52 cards, there are 13 diamonds and 13 spades. ∴ Number of ways of drawing 3 diamonds and one spade $=^{13} C_{3} \times^{13} C_{1}$ Thus, the probability of obtaining 3 diamonds and one spade $= \frac{^{13} C _{3} \times ^{13} C _{1}}{^{52} C _{4}}$
A die has two faces each with number '1', three faces each with number '2' and one face with number ' 3 '. If die is rolled once, determine (i) P(2) (ii) $P (1$ or 3 $)$ (iii) P (not 3 ) Sol. Total number of faces = 6 (i) Number faces with number '2' = 3 $∴ P (2) = \frac{3}{6} = \frac{1}{2}$ (ii) $P (1$ or 3 $) = P ($ not 2 $) = 1 - P (2) = 1 - \frac{1}{2} = \frac{1}{2}$

(iii) Number of faces with number '3' = 1 $∴ P (3) = \frac{1}{6}$ Thus, $P ($ not 3 $) = 1 - P (3) = 1 - \frac{1}{6} = \frac{5}{6}$

In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (i) one ticket (ii) two tickets (iii) 10 tickets? Sol. Total number of tickets sold $= 10, 000$ Number prizes awarded = 10 (i) If we buy one ticket, then

$P (getting a prize) = \frac{10}{10000} = \frac{1}{1000}$

$∴ P (not getting a prize) = 1 - \frac{1}{1000} = \frac{999}{1000}$ (ii) If we buy two tickets, then Number of tickets not awarded

$= 10, 000 - 10 = 9990$

$P ($ not getting a prize $) = \frac{^{9990} C _{2}}{^{10000} C _{2}}$ (iii) If we buy 10 tickets, then

$P (not getting a prize) = \frac{^{9990} C _{10}}{^{10000} C _{10}}$

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same sections? (b) you both enter the different sections? Sol. My friend and I are among the 100 students. Total number of ways of selecting 2 students out of 100 students $=^{100} C_{2}$ (a) The two of us will enter the same section if both of us are among 40 students or among 60 students. $∴$ Number of ways in which both of us enter the same section $=^{40} C_{2} +^{60} C_{2}$ ∴ Probability that both of us enter the same

$section = \frac{^{40} C _{2} + ^{60} C _{2}}{^{100} C _{2}} = \frac{\frac{⌊ 40}{⌊ 2 ⌋ 38} + \frac{⌊ 60}{⌊ 2 ⌋ 58}}{\frac{⌊ 100}{⌊ 2 ⌋ 98}} = \frac{( 30 \times 40 ) + ( 59 \times 60 )}{99 \times 100} = \frac{17}{33}$

(b) $P$ (we enter different sections)

$= 1 - P (we enter the same section) = 1 - \frac{17}{33} = \frac{16}{33}$

Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. Sol. Let $L_{1}, L_{2}, L_{3}$ be three letters and $E_{1}, E_{2}$ , and $E_{3}$ be their corresponding envelops respectively. There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

$L_{1} E_{1}, L_{2} E_{3}, L_{3} E_{2} L_{2} E_{2}, L_{1} E_{3}, L_{3} E_{1} L_{3} E_{3}, L_{1} E_{2}, L_{2} E_{1} L_{1} E_{1}, L_{2} E_{2}, L_{3} E_{3} L_{1} E_{2}, L_{2} E_{3}, L_{3} E_{1} L_{1} E_{3}, L_{2} E_{1}, L_{3} E_{2}$

There are 4 ways in which at least one letter is inserted in a proper envelope. Thus, the required probability is $\frac{4}{6} = \frac{2}{3}$

A and B are two events such that $P (A) = 0.54$ , $P (B) = 0.69$ and $P (A \cap B) = 0.35$ . Find (i) $P (A \cap B)$ (ii) $P (A^{'} \cap B^{'})$ (iii) $P (A \cap B^{'})$ (iv) $P (B \cap A^{'})$

Sol. It is given that

$P (A) = 0.54, P (B) = 0.69, P (A \cap B) = 0.35$

(i) We know that

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

(ii) $A^{'} \cap B^{'} = (A \cup B)^{'} [$ by De Morgan's law]

$∴ P (A^{'} \cap B^{'}) = P (A \cup B)^{'} = 1 - P (A \cup B) = 1 - 0.88 = 0.12$

(iii) $P (A \cap B^{'}) = P (A) - P (A \cap B)$

$= 0.54 - 0.35 = 0.19$

(iv) We know that :

$\Rightarrow n (B \cap A^{'}) = n (B) - n (A \cap B) \frac{n ( B \cap A ^{'} )}{n ( S )} = \frac{n ( B )}{n ( S )} - \frac{n ( A \cap B )}{n ( S )}$

∴ P(B ∩ A') = P(B) – P(A ∩ B) ∴ P(B ∩ A') = 0.69 – 0.35 = 0.34

From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

S. No.	Name	Sex	Age in years
1.	Harish	M	30
2.	Rohan	M	33
3.	Sheetal	F	46
4.	Alis	F	28
5.	Salim	M	41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? Sol. Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age. Accordingly, $P (E) = 3/5$ and $P (F) = 2/5$ Since there is only one male who is over 35 years of age,

$P (E \cap F) = \frac{1}{5}$

We know that:

$∴ P (E \cup F) = P (E) + P (F) - P (E \cap F) P (E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5}$

Thus, the probability that the spokesperson will either be a male or over 35 years of age is 4/5.

If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed? Sol. (i) When the digits are repeated Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5. The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed. ∴ Total number of 4-digit numbers greater than $5000 = 2 \times 5 \times 5 \times 5 - 1 = 250 - 1 = 249$ [In this case, 5000 cannot be counted; so 1 is subtracted] A number is divisible by 5 if the digit at its units place is either 0 or 5. ∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5

$= 2 \times 5 \times 5 \times 2 - 1 = 100 - 1 = 99$

Thus, the probability of forming a number divisible by 5 when the digits are repeated is $= \frac{99}{249} = \frac{33}{83}$ (ii) When repetition of digits is not allowed The thousands place can be filled with either of the two digits 5 or 7. The remaining 3 places can be filled with any of the remaining 4 digits. ∴ Total number of 4-digit numbers greater than $5000 = 2 \times 4 \times 3 \times 2 = 48$ When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.

$∴$ Here, number of 4-digit numbers starting with 5 and divisible by $5 = 3 \times 2 = 6$ When the digit at the thousands place is 7 , the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits. $∴$ Here, number of 4-digit numbers starting with 7 and divisible by $5 = 1 \times 2 \times 3 \times 2 = 12$ ∴ Total number of 4-digit numbers greater than 5000 that are divisible by $5 = 6 + 12 = 18$ Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is $\frac{18}{48} = \frac{3}{8}$

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase? Sol. The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9. Number of ways of selecting 4 different digits out of the 10 digits $=^{10} C_{4}$ Now, each combination of 4 different digits can be arranged in 4! Ways. ∴ Number of four digits with no repetitions $= 4!^{10} C_{4} = 5040$ There is only one number that can open the suitcase. Thus, the required probability is $\frac{1}{5040}$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Access chapter-wise NCERT Class 11 Maths solutions with accurate answers, important formulas, and clear explanations. Understand every concept, solve NCERT questions step by step, and strengthen your problem-solving skills for exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 14

Exercise	Questions & Solutions	Important Concepts Covered
Exercise 14.1	7 Questions & Solutions	Probability, types of events, simple events, compound events, mutually exclusive events, independent events
Exercise 14.2	21 Questions & Solutions	Addition rule, multiplication rule, and probability-based problems
Miscellaneous Exercise	10 Questions & Solutions	Addition rule, multiplication rule, conditional probability, Bayes’ theorem, and mixed probability problems

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 14 (Probability)

Clear Understanding of Events and Sets: Learn how to represent events using set notation and understand the relationship between sample space, events, and outcomes.
Step-by-Step Sample Space Construction: Understand how to list possible outcomes for experiments such as tossing coins, rolling dice, and drawing balls from an urn. Tree diagrams can also be used to organise complex outcomes clearly.
Use of Permutations and Combinations: Learn how Permutations and Combinations can be used to count total outcomes and favourable outcomes in probability questions.
Understanding Probability Statements: Learn how to solve questions involving terms such as at least one, exactly one, and not more than by converting them into clear mathematical conditions.
Probability of Events: Understand how to calculate the probability of an event using the number of favourable outcomes and the total number of outcomes.
Prepared by ALLEN Subject Experts:
These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Clear Step-by-Step Presentation: Each solution uses simple notation and logical steps, making probability questions easier to understand and solve correctly.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with clear solutions to help students understand and apply the concepts of Probability.

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Why are NCERT Solutions for Class 11 Maths Chapter 14 Probability important for students?

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What concepts are covered in NCERT Solutions for Class 11 Maths Chapter 14 Probability?

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Frequently Asked Questions

Why are NCERT Solutions for Class 11 Maths Chapter 14 Probability important for students?

How does Class 11 Maths NCERT Solutions Chapter 14 help in Board and JEE prep?

What concepts are covered in NCERT Solutions for Class 11 Maths Chapter 14 Probability?

Are NCERT Solutions for Class 11 Maths Chapter 14 by ALLEN useful for competitive exams?

Why is probability important in mathematics and real life?

What is a sample space in Class 11 Maths Chapter 14?

Join ALLEN!

NCERT Solutions Class 11 Maths Chapter 14 – Probability

1.0Class 11 Maths Chapter 14: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 8

EXERCISE - 14.1

EXERCISE - 14.2

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 14

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 14 (Probability)

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