NCERT Solutions for Class 11 Maths Chapter 14  Probability
Probability Class 11 chapter is the one chapter in which students are confident in answering all the questions. It is one of the easiest topics in all the subjects that Class 11th offers to students. There are several benefits of learning Probability to the core as it can help you in daily life in making complex decisions. Probability might be the most manageable topic in the Class 11th Maths textbook, but it is one of the topics you will use throughout your life.
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Chapter Name:  Chapter 14  Probability 
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From weather reports to fiscal reports, all of them use Probability in one way or the other. It's a way to find out a little bit about the future based on the information which you have in the present.
Probability Chapter at a Glance  Class 11 NCERT Solutions
Sample space: The set of all possible outcomes.
Sample points: Elements of sample space
Event: A subset of the sample space
Impossible event: The empty set
Sure event: The whole sample space
Complementary event or 'not event': The set $A^{\prime}$ or $SA$
Event $A$ or $B$ : The set $A \cup B$
Event $A$ and $B$ : The set $A \cap B$
Event $A$ and not $B$ : The set $AB$
Mutually exclusive event: $A$ and $B$ are mutually exclusive if $A \cap B=\phi$
Exhaustive and mutually exclusive events: Events $E_1, E_2, \ldots, E_n$ are mutually exclusive and exhaustive if
$$ \begin{aligned} & E_1 \cup E_2 \cup \ldots \cup E_n=S \text { and } \\ & E_i \cap E_j=\phi \forall i \neq j \end{aligned} $$
Probability: Number $P\left(\omega_i\right)$ associated with sample point $\omega_i$ such that
(i) $0 \leq P\left(\omega_i\right) \leq 1$
(ii) $\sum P\left(\omega_i\right)$ for all $\omega_i \in S=1$
(iii) $P(A)=\sum P\left(\omega_i\right)$ for all $\omega_i \in A$. The number $P\left(\omega_i\right)$ is called probability of the outcome $\omega_i$.
Equally likely outcomes: All outcomes with equal probability.
Probability of an event: For a finite sample space with equally likely outcomes,
Probability of an event $P(A)=\frac{n(A)}{n(S)}$, where $n(A)=$ number of elements in the set $A, n(S)=$ number of elements in the set $S$.
If $A$ and $B$ are any two events, then $P(A$ or $B)=P(A)+P(B)P(A$ and $B)$ equivalently, $P(A \cup B)=P(A)+P(B)P(A \cap B)$
If $A$ and $B$ are mutually exclusive, then $P(A$ or $B)=P(A)+P(B)$
If $A$ is any event, then $P($ not $A)=1P(A)$
Exercises under NCERT Class 11 Maths Chapter 14 Probability
Exercise 14.1: This exercise introduces the concept of probability and its importance in reallife applications. Students will learn about the definition of probability and different types of events, including simple events, compound events, mutually exclusive events, and independent events.
Exercise 14.2: In this exercise, students will learn about the rules of probability, including the addition rule and the multiplication rule. They will also practice solving problems related to probability using these rules.
Exercise 14.3: This exercise focuses on conditional probability and Bayes' theorem. Students will learn about the definition of conditional probability, the multiplication rule of probability, and Bayes' theorem. They will also practice solving problems related to conditional probability and Bayes' theorem.
Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of probability to solve various problems and answer questions. They will also practice using the addition rule, multiplication rule, conditional probability, and Bayes' theorem to solve probability problems.
Access NCERT Solutions for Class11 Maths Chapter 14 Probability
Exercise 14.1
1. Describe the sample space for the indicated experiment: A coin is tossed three times.
Ans. Sample space of an experiment refers to the set of all possible outcomes of that experiment.
Every coin has two faces: Head (H) and Tail (T).
When a coin is tossed thrice, the total number of possible outcomes is \[{2^3} = 8\] .
Thus, when a coin is tossed three times, sample space is obtained as:
\[S = \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \]
2. Describe the sample space for the indicated experiment: A die is thrown two times.
Ans: When a die is thrown, the possible outcomes are \[1,2,3,4,5,6\] .
When a die is thrown twice, the sample space will be obtained as \[S = \{ (x,y):x,y = 1,2,3,4,5,6\} \] .
Total elements in this sample space are $6 \times 6 = 36$, while the sample space is given by:
$S = \ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\$
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\$
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\$
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),\\$
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\ $
3. Describe the sample space for the indicated experiment: A coin is tossed four times.
Ans. Every coin has two faces: Head (H) and Tail (T).
When a coin is tossed four times, the total number of possible outcomes is \[{2^4} = 16\] .
When a coin is tossed four times, the sample space is obtained as:
S = HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,
THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT
4. Describe the sample for the indicated experiment: A coin is tossed and a die is thrown.
Ans. A coin has two faces: Head (H) and Tail (T).
A die has six faces numbered from \[1\] to \[6\] , with one number on each face. Thus, when a coin is tossed and a die is thrown simultaneously, the sample space is obtained as:
\[S = \{ (H1),\;(H2),\;(H3),\;(H4),\;(H5),\;(H6),(T1),(T2),(T3),(T4),(T5),(T6)\} \]
5. Describe the sample space for the indicated experiment: A coin is tossed and then a die is rolled only in case a head is shown on the coin.
Ans. A coin has two faces: Head (H) and Tail (T).
A die has six faces numbered from \[1\] to \[6\] , with one number on each face.
Thus, when a coin is tossed and a die is rolled only if a head is shown on the coin, the sample space is obtained as:
\[S = \{ (H1),\;(H2),\;(H3),\;(H4),\;(H5),\;(H6),(T)\} \]
6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person
Ans. We denote 2 boys and 2 girls in Room X as $B_1 , B_2 , G_1 , G_2$ respectively and 1 boy and 3 girls in Room Y as $B_3,G_3,G_4,G_5$ respectively.
Consequently, the required sample space is obtained as:
\[S = \{ {\rm{X}}{{\rm{B}}_1}{\rm{,X}}{{\rm{B}}_2}{\rm{,X}}{{\rm{G}}_{\rm{1}}}{\rm{,X}}{{\rm{G}}_{\rm{2}}}{\rm{,Y}}{{\rm{B}}_{\rm{3}}}{\rm{,Y}}{{\rm{G}}_{\rm{3}}}{\rm{, Y}}{{\rm{G}}_{\rm{4}}}{\rm{, Y}}{{\rm{G}}_{\rm{5}}}\} \]
7. One die of red color, one of white color and one of blue color are placed in a bag. One die is selected at random and rolled, its color and the number on its uppermost face is noted. Describe the sample space.
Ans. A die has six faces that are numbered from \[1\] to \[6\] . We denote the red, white, and blue dices as R, W, and B respectively.
Consequently, when a die is selected and then rolled, the sample is obtained as
\[S = \{ R1,R2,R3,R4,R5,R6,W1,W2,W3,W4,W5,W6,B1,B2,B3,B4,B5,B6\} \]
8. An experiment consists of recording boygirl composition of families with \[2\] children.
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
Ans (i). When the order of the birth of a girl or a boy has to be considered, the sample space will be obtained as:
\[S = \{ GG,GB,BG,BB\} \]
(ii) What is the sample space if we are interested in the number of girls in the family?
Ans (ii). Since the maximum number of children in each family is \[2\] , a family can possibly have \[2\] girls or \[1\] girl or no girl.
So, the required sample space is:
\[S = \{ 0,1,2\} \]
9. A box contains \[1\] red and \[3\] identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Ans. It is given that a box contains \[1\] red ball and \[3\] identical white balls. We denote the red ball with R and white ball with W.
When two balls are drawn randomly in succession without replacement, the sample space is obtained as:
\[S = \{ RW,WR,{\rm{ }}WW\} \]
10. An experiment consists of tossing a coin and then throwing it a second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.
Ans. A coin has two faces: Head (H) and Tail (T).
A die has six faces that are numbered from \[1\] to \[6\] , with one number on each face. The experiment states that if a head occurs on tossing the coin, the it will be tossed a second time. However, if a tail occurs on tossing, then a die will be rolled.
Thus, in the given experiment, the sample space is obtained as:
\[S = \{ HH,HT,T1,T2,T3,T4,T5,T6\} \]
11. Suppose \[3\] bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or nondefective (N). Write the sample space of this experiment?
Ans. \[3\] bulbs are selected randomly from a given lot. Further, each bulb is tested and then classified as defective (D) or nondefective (N).
The sample space of this experiment is obtained as:
\[S = \{ DDD,DND,DNN,NDD,NDN,NND,NNN\} \]
12. A coin is tossed. If the outcome is the head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?
Ans. On tossing a coin, possible outcomes are Head (H) or Tail (T).
On throwing a die, the possible outcomes are \[1,2,3,4,5,6\] .
Thus, the sample space of this experiment is obtained as:
\[\begin{array}{l}S = \{ T,H1,H3,H5,H21,H22,H23,H24,H25,H26,\\H41,H42,H43,H44,H45,H46,H61,H62,H63,H64,H65,H66\} \end{array}\]
13. The numbers \[1\] , \[2\] , \[3\] and \[4\] are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Ans. If \[1\] appears on the first drawn slip, then the possible numbers to appears on the second drawn slip are \[2\] , \[3\] or \[4\] . Similarly, if \[2\] appears on the first drawn slip, then the possible numbers to appears on the second drawn slip are \[1\] , \[3\] or \[4\] . The same holds true for the remaining numbers also.
Thus, the sample space of this experiment is obtained as:\[S = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {2,1} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,4} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right)} \right\}\]
14. An experiment consists of rolling a die and then tossing a coin if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.
Ans. A die has six faces numbered from \[1\] to \[6\] . Among these numbers, \[2\] , \[4\] , and \[6\] are even, while \[1\] , \[3\] , and \[5\] are odd.
There are two faces of a coin: Head (H) and Tail (T).
Thus, the sample space of this experiment is obtained as:
\[\begin{array}{l}S = \{ 2H,2T,4H,4T,6H,6T,1HH,1HT,1TH,1TT,3HH,3HT,3TH,3TT,\\5HH,5HT,5TH,5TT\} \end{array}\]
15. A coin is tossed. If it shows a tail, we draw a ball from a box which contains \[2\] red and \[3\] black balls. If it shows head, we throw a die. Find the sample space for this experiment.
Ans. The box contains \[2\] red balls and \[3\] black balls. We denote the \[2\] red balls as \[{R_1}\] , \[{R_2}\] and the \[3\] black balls as \[{B_1}\] , \[{B_2}\] and \[{B_3}\] .
Thus, the sample space of this experiment is given as:
\[S = \{ T{R_1},T{R_2}\;,T{B_1},T{B_2}\;,T{B_3},H1,H2,H3,H4,H5,H6\} \]
16. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?
Ans. In this experiment, six might come up on the first throw, or the second throw, or the third throw and so on till six is obtained.
Hence, the sample space of this experiment is obtained as:
\[\begin{array}{l}S = \{ 6,\left( {1,6} \right),\left( {2,6} \right),\left( {3,6} \right),\left( {4,6} \right),\left( {5,6} \right),\\\left( {1,1,6} \right),\left( {1,2,6} \right), \ldots ,\left( {1,5,6} \right),\left( {2,1,6} \right),\left( {2,2,6} \right), \ldots ,\left( {2,5,6} \right),\\\left( {3,1,6} \right),\left( {3,2,6} \right), \ldots ,\left( {3,5,6} \right),\left( {4,1,6} \right),\left( {4,2,6} \right), \ldots ,\left( {4,5,6} \right),\\\left( {5,1,6} \right),\left( {5,2,6} \right), \ldots ,(5,5,6)...\} \end{array}\]
Exercise 14.2
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?
Ans. When a dice is rolled, the sample space is \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
Further, \[E = \left\{ 4 \right\}\] and \[F = \left\{ {2,4,6} \right\}\] .
It is observed that \[E \cap F = \left\{ 4 \right\}\] .
Hence, E and F are not mutually exclusive events.
2. A die is thrown. Describe the following events:
(i) A: a number less than \[7\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[A = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\]
(ii) B: a number greater than \[7\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[B = \{ \Phi \} \]
(iii) C: a multiple of \[3\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[C = \left\{ {3,{\bf{6}}} \right\}\]
(iv) D: a number less than \[4\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[D = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}}} \right\}\]
(v) E: an even number greater than \[4\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[E = \left\{ {\bf{6}} \right\}\]
(vi) F: a number not less than \[3\]
Ans. When a die is rolled, the sample space is obtained as \[S = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] .
\[F = \left\{ {3,4,5,{\bf{6}}} \right\}\]
(vii) Also find \[A \cup B\] , \[A \cap B\] , \[B \cup C\] , \[E \cap F\] , \[D \cap E\] , \[A  C\] , \[D  E\] , \[E \cap F'\] and \[F'\] .
Ans. We know that \[A = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\] , \[B = \{ \Phi \} \] , \[C = \left\{ {3,{\bf{6}}} \right\}\] , \[D = \left\{ {{\bf{1}},{\bf{2}},{\bf{3}}} \right\}\] , \[E = \left\{ {\bf{6}} \right\}\] , \[F = \left\{ {3,4,5,{\bf{6}}} \right\}\] . Further, \[F' = \left\{ {1,2} \right\}\] .
So, \[A \cup B = \left\{ {1,2,3,4,5,{\bf{6}}} \right\}\]
\[A \cap B = \left\{ \Phi \right\}\]
\[B \cup C = \left\{ {3,{\bf{6}}} \right\}\]
\[E \cap F = \left\{ 6 \right\}\]
\[D \cap E = \left\{ \Phi \right\}\]
\[A  C = \left\{ {1,2,4,5} \right\}\]
\[D  E = \left\{ {1,2,3} \right\}\]
\[E \cap F' = \left\{ \Phi \right\}\]
3. An experiment involves rolling a pair of dice and recording the number that comes 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 multiple of \[3\] .
Which pairs of these events are mutually exclusive?
Ans. On rolling a pair of dice, the sample space is obtained as: \[S = \left\{ {\left( {{\bf{x}},{\bf{y}}} \right):{\bf{x}},{\bf{y}} = {\bf{1}},{\bf{2}},{\bf{3}},{\bf{4}},{\bf{5}},{\bf{6}}} \right\}\]
\[ = \left\{ \begin{array}{l}\left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\]
Subsequently,
\[A = \{ \left( {3,6} \right),\left( {4,5} \right),\left( {4,6} \right),(5,\;4),\left( {5,5} \right),\left( {5,6} \right)\left( {6,3} \right),(6,\;4),\left( {6,5} \right),\left( {6,6} \right)\} \]
\[B = \{ \left( {2,1} \right),(2,\;2),\left( {2,3} \right),(2,\;4),\left( {2,5} \right),\left( {2,6} \right)(1,\;2),(3,\;2),(4,\;2),(5,\;2),(6,\;2)\} \]
\[C = \{ \left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right),\left( {6,6} \right)\} \]
It is further observed that:
\[A \cap B = \left\{ \Phi \right\}\]
\[B \cap C = \left\{ \Phi \right\}\]
\[A \cap C = \left\{ {\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right),\left( {6,6} \right)} \right\}\]
Thus, events A and B and events B and C are mutually exclusive.
4. 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?
Ans. On rolling a pair of dice, the sample space is obtained as:
\[S = \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \]
Further, \[A = \{ HHH\} \] , \[B = \{ HHT,HTH,THH\} \] , \[C = \{ TTT\} \] , \[D = \{ HHH,HHT,HTH,HTT\} \] .
We conclude that \[A \cap B = \left\{ \Phi \right\}\] , \[A \cap C = \left\{ \Phi \right\}\] , \[A \cap D = \left\{ {HHH} \right\}\] , \[B \cap C = \left\{ \Phi \right\}\] , \[B \cap D = \left\{ {HHT,HTH} \right\}\] , \[C \cap D = \left\{ \Phi \right\}\] .
Hence, events A and B; A and C; B and C; and C and D are all mutually exclusive.
(ii) simple?
Ans. A simple event is an event that has only one sample point of a sample space. Therefore, A and C are simple events.
(iii) compound?
Ans. A compound event is an event that has more than one sample point of a sample space. Therefore, B and D are compound events.
5. Three coins are tossed. Describe:
(i) Two events which are mutually exclusive.
Ans. On rolling a pair of dice, the sample space is obtained as:
\[S = \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \] .
Two mutually exclusive events can be:
A: Getting no heads, and B: Getting no tails.
Here, \[A = \{ TTT\} \] and \[B = \{ HHH\} \] are disjoint.
(ii) Three events which are mutually exclusive and exhaustive.
Ans. Three mutually exclusive events are:
A: Getting no heads, B: Getting exactly one head, and C: Getting at least two heads.
Here, \[A = \{ TTT\} \] , \[B = \{ HTT,THT,TTH\} \] , \[C = \{ HHH,HHT,HTH,THH\} \] are disjoint but \[A \cup B \cup C = S\] .
(iii) Two events, which are not mutually exclusive.
Ans. Two events that are not mutually exclusive are:
A: Getting three heads, and B: Getting at least two heads.
Here, \[A = \{ HHH\} \] , \[B = \{ HHH,HHT,HTH,THH\} \] and \[A \cap B = \{ HHH\} \] .
(iv) Two events which are mutually exclusive but not exhaustive.
Ans. Two mutually exclusive but not exhaustive events are:
A: Getting exactly one head, and B: Getting exactly one tail.
Here, \[A = \{ HTT,THT,TTH\} \] , \[B = \{ HHT,HTH,THH\} \] are disjoint and \[A \cup B \ne S\] .
(v) Three events which are mutually exclusive but not exhaustive.
Ans. Three mutually exclusive but not exhaustive events are:
A: Getting exactly three heads, B: Getting one tail and two heads, C: Getting one head and two tails.
Here, \[A = \{ HHH\} \] , \[B = \{ HTH,HHT,THH\} \] , \[C = \{ HTT,TTH,THT\} \] are disjoint and \[A \cup B \cup C \ne S\]
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 number on the dice \[ \le 5\] .
Describe the events:
(i) \[A'\]
Ans. When two dice are thrown, the sample space is obtained as:
\[S = \{ (x,y):x,y = 1,2,3,4,5,6\} \]\[ \Rightarrow \left\{ \begin{array}{l}\left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\]Subsequently,
\[\begin{array}{l} A = \{ \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right),\\\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\} \end{array}\]\
\[\begin{array}{l}B = \{ \left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {3,1} \right),(3,2)(3,3),(3,4),(3,5),(3,6),\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6)\} \end{array}\]\[C = \{ \left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),(2,1),(2,2),(2,3),\;\left( {3,1} \right),(3,2),(4,1)\} \]
Further, \[\begin{array}{l}A' = \{ \left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6)\} \end{array}\]
\[ \Rightarrow A' = B\]
(ii) not B
Ans. Not B can be rewritten as \[B'\] .
\[\begin{array}{l}B' = \{ \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right),\\\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\} \end{array}\]
\[ \Rightarrow B' = A\]
(iii) A or B
Ans. A or B can be written as \[A \cup B\] .
\[A \cup B = \left\{ \begin{array}{l}\left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\]
\[A \cup B = S\]
(iv) A and B
Ans. A and B, or \[A \cap B\] is \[\Phi \] .
(v) A but not C
Ans. A but not C, or \[A  C\] ,
\[\begin{array}{l}A  C = \{ \left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\\(4,6),(6,1),(6,2),(6,3),(6,4),(6,5)(6,6)\} \end{array}\]
(vi) B or C
Ans. B or C, or \[B \cup C\] , \[\begin{array}{l}B \cup C = \{ {\rm{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),}}\\{\rm{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} }}\end{array}\]
(vii) B and C
Ans. B and C, or \[B \cap C\] ,
\[B \cap C = {\rm{\{ (1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} }}\]
(viii) \[A \cap B' \cap C'\]
Ans. We comprehend $C^{‘}$as follows:
\[\begin{array}{l}C' = \{ {\rm{(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),}}\\{\rm{(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} }}\end{array}\]
Now, \[A \cap B' \cap C' = A \cap A \cap C'\]
\[ \Rightarrow A \cap C'\]\[\begin{array}{l}\Rightarrow A \cap C' = \{ (2,4),(2,5),(2,6)(4,3),(4,4),(4,5),(4,6),\\(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{array}\]
7. 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 number on the dice \[ \le 5\] .
State true or false (give reason for your answer) :
(i) A and B are mutually exclusive.
Ans. When two dice are thrown, the sample space is obtained as:
\[S = \{ (x,y):x,y = 1,2,3,4,5,6\} \]\[ \Rightarrow \left\{ \begin{array}{l}\left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\\\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\]
Accordingly,
\[\begin{array}{l}A = \{ \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right),\\\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\} \end{array}\]
\[\begin{array}{l}B = \{ \left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),\;(1,\;5),\;(1,\;6),\left( {3,1} \right),(3,2),(3,3),(3,4),(3,5),(3,6),\\\left( {5,1} \right),(5,2),(5,3),(5,4),(5,5),(5,6)\} \end{array}\]\[C = \{ \left( {1,1} \right),\;(1,\;2),\;(1,\;3),\;(1,\;4),(2,1),(2,2),(2,3),\;\left( {3,1} \right),(3,2),(4,1)\} \]
So, \[A \cap B = \{ \Phi \} \] , that is, the sets are mutually exclusive.
Hence the statement is true.
(ii) A and B are mutually exclusive and exhaustive.
Ans. We observe that \[A \cap B = \{ \Phi \} \] , that is, the sets are mutually exclusive. Further, \[A \cup B = S\] , so the sets are also exhaustive.
Hence the statement is true.
(iii) \[A = B'\]
Ans. We observe
\[\begin{array}{l}B' = \{ \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,1} \right),\left( {4,2} \right),\\\left( {4,3} \right),(4,4),\left( {4,5} \right),(4,6).\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\} \end{array}\] , that is, \[A = B'\] .
Hence the statement is true.
(iv) A and C are mutually exclusive.
Ans. We observe that \[A \cap B = \{ (2,1),(2,2),(2,3),(4,1)\} \] .
Thus, A and C are not mutually exclusive.
Hence the statement is false.
(v) A and \[B'\] are mutually exclusive.
Ans. We observe
\[\begin{array}{l}B' = \{ \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,1} \right),\left( {4,2} \right),\\\left( {4,3} \right),(4,4),\left( {4,5} \right),(4,6).\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\} \end{array}\] , that is, \[A = B'\] .
So, A and \[B'\] are not mutually exhaustive.
Hence the given statement is false.
(vi) \[A'\] , \[B'\] , C are mutually exclusive and exhaustive.
Ans. We observe that \[A' \cup B' \cup C = S\] , which means they are exhaustive.
However, \[B' \cap C \ne \Phi \] , so they are not mutually exhaustive.
Hence the given statement is false.
Exercise 14.3
1. Which of the following cannot be valid assignment of probabilities for outcomes of sample space \[S = \{ {\omega _1},{\omega _2},{\omega _3},{\omega _4},{\omega _5},{\omega _6},{\omega _7}\} \]
Assignment  \[{\omega _1}\]  \[{\omega _2}\]  \[{\omega _3}\]  \[{\omega _4}\]  \[{\omega _5}\]  \[{\omega _6}\]  \[{\omega _7}\] 
(a)  \[0.1\]  \[0.01\]  \[0.05\]  \[0.03\]  \[0.01\]  \[0.2\]  \[0.6\] 
Ans. Here, each of the given probabilities is less than \[1\] and positive.
Further, sum of possibilities:
\[{\omega _1} + {\omega _2} + {\omega _3} + {\omega _4} + {\omega _5} + {\omega _6} + {\omega _7}\]
\[ \Rightarrow 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6\]
\[ \Rightarrow 1\]
Hence, the given assignment is valid.
Assignment  \[{\omega _1}\]  \[{\omega _2}\]  \[{\omega _3}\]  \[{\omega _4}\]  \[{\omega _5}\]  \[{\omega _6}\]  \[{\omega _7}\] 
(b)  \[\frac{1}{7}\]  \[\frac{1}{7}\]  \[\frac{1}{7}\]  \[\frac{1}{7}\]  \[\frac{1}{7}\]  \[\frac{1}{7}\]  \[\frac{1}{7}\] 
Ans. Here, each of the given probabilities is less than \[1\] and positive.
Further, sum of possibilities:
\[{\omega _1} + {\omega _2} + {\omega _3} + {\omega _4} + {\omega _5} + {\omega _6} + {\omega _7}\]
\[ \Rightarrow \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}\]
\[ \Rightarrow 1\]
Hence, the given assignment is valid.
Assignment  \[{\omega _1}\]  \[{\omega _2}\]  \[{\omega _3}\]  \[{\omega _4}\]  \[{\omega _5}\]  \[{\omega _6}\]  \[{\omega _7}\] 
(c)  \[0.1\]  \[0.2\]  \[0.3\]  \[0.4\]  \[0.5\]  \[0.6\]  \[0.7\] 
Ans. Here, each of the given probabilities is less than \[1\] and positive.
Further, sum of possibilities:
\[{\omega _1} + {\omega _2} + {\omega _3} + {\omega _4} + {\omega _5} + {\omega _6} + {\omega _7}\]
\[ \Rightarrow 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7\]
\[ \Rightarrow 2.8\]
Since \[2.8 > 1\] , the given assignment is not valid.
Assignment  \[{\omega _1}\]  \[{\omega _2}\]  \[{\omega _3}\]  \[{\omega _4}\]  \[{\omega _5}\]  \[{\omega _6}\]  \[{\omega _7}\] 
(d)  \[  0.1\]  \[0.2\]  \[0.3\]  \[0.4\]  \[  0.2\]  \[0.1\]  \[0.3\] 
Ans. Here, \[{\omega _1}\] and \[{\omega _5}\] are less than \[1\] and negative.
Hence, the given assignment is not valid.
Assignment  \[{\omega _1}\]  \[{\omega _2}\]  \[{\omega _3}\]  \[{\omega _4}\]  \[{\omega _5}\]  \[{\omega _6}\]  \[{\omega _7}\] 
(e)  \[\frac{1}{{14}}\]  \[\frac{2}{{14}}\]  \[\frac{3}{{14}}\]  \[\frac{4}{{14}}\]  \[\frac{5}{{14}}\]  \[\frac{6}{{14}}\]  \[\frac{{15}}{{14}}\] 
Ans. Here, \[{\omega _7}\] is more than \[1\].
Hence, the given assignment is not valid.
2. A coin is tossed twice, what is the probability that at least one tail occurs?
Ans. When a coin is tossed twice, the sample space is obtained as:
\[S = \{ {\bf{HH}},{\bf{HT}},{\bf{TH}},{\bf{TT}}\} \]
Let A be the event that tail occurs at least once.
Accordingly, \[{\bf{A}} = \left\{ {{\bf{HT}},{\bf{TH}},{\bf{TT}}} \right\}\]
So, the probability of A occurring is the number of favorable outcomes over the total possible outcomes.
\[P({\bf{A}}) = \frac{{n(A)}}{{n(S)}}\]
\[P({\bf{A}}) = \frac{3}{4}\]
3. A die is thrown, find the probability of following events:
(i) A prime number will appear,
Ans. When a die is thrown, the sample space is obtained as:
\[S = \{ 1,2,3,4,5,6\} \]
Let A be the event of a prime number occurring. So, \[A = \{ 2,3,5\} \] .
\[P({\bf{A}}) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow P({\bf{A}}) = \frac{3}{6}\]
\[ \Rightarrow P({\bf{A}}) = \frac{1}{2}\]
(ii) A number greater than or equal to \[3\] will appear,
Ans. When a die is thrown, the sample space is obtained as:
\[S = \{ 1,2,3,4,5,6\} \]
Let B be the event a number greater than or equal to \[3\] occurs. So, \[B = \{ 3,4,5,6\} \].
\[P(B) = \frac{{n(B)}}{{n(S)}}\]
\[ \Rightarrow P(B) = \frac{4}{6}\]
\[ \Rightarrow P(B) = \frac{2}{3}\]
(iii) A number less than or equal to \[1\] will appear,
Ans. When a die is thrown, the sample space is obtained as:
\[S = \{ 1,2,3,4,5,6\} \]
Let C be the event a number less than or equal to \[1\] occurs. So, \[C = \{ 1\} \] .
\[P(C) = \frac{{n(C)}}{{n(S)}}\]
\[ \Rightarrow P(C) = \frac{1}{6}\]
(iv) A number more than \[6\] will appear,
Ans. Let D be the event that a number greater than six appears.
So, \[D = \Phi \] .
\[P(D) = \frac{{n(D)}}{{n(S)}}\]
\[ \Rightarrow P(D) = \frac{0}{6}\]
\[ \Rightarrow P(D) = 0\]
(v) A number less than \[6\] will appear.
Ans. Let E be the event that a number less than six appears.
So, \[E = \{ 1,2,3,4,5\} \] .
\[P(E) = \frac{{n(E)}}{{n(S)}}\]
\[ \Rightarrow P(E) = \frac{5}{6}\]
4. A card is selected from a pack of \[52\] cards.
(i) How many points are there in the sample space?
Ans. When a card is selected from a pack \[52\] cards, the number of possible outcomes is \[52\] , that is, the sample space contains \[52\] elements.
Hence, there are \[52\] points in the sample space.
(ii) Calculate the probability that the card is an ace of spades.
Ans. Let A be the event in which the card drawn is an ace of spades. Subsequently, \[n(A) = 1\] .
\[P({\bf{A}}) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow P({\bf{A}}) = \frac{1}{{52}}\]
(iii) Calculate the probability that the card is:
an ace
Ans. Let E be the event in which the card drawn is an ace. Subsequently, \[n(E) = 4\] .
\[P(E) = \frac{{n(E)}}{{n(S)}}\]
\[ \Rightarrow P(E) = \frac{4}{{52}}\]
\[ \Rightarrow P(E) = \frac{1}{{13}}\]
black card.
Ans. Let F be the event in which the card drawn is black. Subsequently, \[n(F) = 26\] .
\[P(F) = \frac{{n(F)}}{{n(S)}}\]
\[ \Rightarrow P(F) = \frac{{26}}{{52}}\]
\[ \Rightarrow P(F) = \frac{1}{2}\]
5. 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) \[3\]
Ans. Since the fair coin has \[1\] marked on one face and \[6\] on the other, and the die has six faces that are numbered \[1,2,3,4,5,6\] , the sample space is obtained as:
\[S = \{ (1,1),(1,2),(1,3)(1,4),(1,5),(1,6),(6,1),(6,2),(6,3)(6,4),(6,5),(6,6)\} \]
Let A be the event in which the sum of numbers that turn up is \[3\] . So, \[A = \{ (1,2)\} \] .
\[P(A) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow P(A) = \frac{1}{{12}}\]
(ii) \[12\]
Ans. Let B be the event in which the sum of numbers that turn up is \[12\] . So, \[B = \{ (6,6)\} \] .
\[P(B) = \frac{{n(B)}}{{n(S)}}\]
\[ \Rightarrow P(B) = \frac{1}{{12}}\]
6. 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?
Ans. There are four men and six women on the city council.
Since one council member has to be selected for the committee randomly, the sample space will contain \[10\] elements.
Let A be the event in which the council member selected is a woman.
Accordingly, \[n(A) = 6\]
\[P(A) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow P(A) = \frac{6}{{10}}\]
\[ \Rightarrow P(A) = \frac{3}{5}\]
7. A fair coin is tossed four times, and a person win Re \[1\] for each head and loss 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
Ans. As the coin is tossed four times, there can be a maximum of \[4\] heads and \[4\] tails:
When \[4\] heads turn up, Rs. \[(1 + 1 + 1 + 1) = \] Rs. \[4\] gain.
When \[3\] heads and \[1\] tail turn up, Rs. \[(1 + 1 + 1  1.50) = \] Rs. \[1.50\] gain.
When \[2\] heads and \[2\] tails turn up, Rs. \[(1 + 1  1.50  1.50) = \] Rs. \[1\] loss.
When \[1\] head and \[3\] tails turn up, Rs. \[(1  1.50  1.50  1.50) = \] Rs. \[3.50\] loss.
When \[4\] tails turn up, Rs. \[(  1.50  1.50  1.50  1.50) = \] Rs. \[6\] loss.
There are \[{2^4} = 16\] elements in the sample space S, which is obtained as:
\[S{\rm{ }} = {\rm{ }}\left\{ \begin{array}{l}HHHH,HHHT,HHTH,HTHH,THHH,HHTT,HTTH,TTHH,{\rm{ }}\\HTHT,THTH,THHT,HTTT,THTT,TTHT,TTHT,TTTH,TTTT\end{array} \right\}\]
\[n(S) = 16\]
The person wins Rs. \[4\] when \[4\] heads turn up, that is, when the event \[\{ HHHH\} \] occurs.
Probability of winning Rs. \[4\] is \[\frac{1}{{16}}\] .
The person wins Rs. \[1.50\] when \[3\] heads and \[1\] tail turn up, that is, when the events \[\left\{ {HHHT,HHTH,HTHH,THHH} \right\}\] occur.
Probability of winning Rs. \[1.50\] is \[\frac{4}{{16}} = \frac{1}{4}\] .
The person loses Rs. \[1\] when \[2\] heads and \[2\] tails turn up, that is, when the events \[\left\{ {HHTT,HTTH,TTHH,HTHT,THTH,THHT} \right\}\] occur.
Probability of losing Rs. \[1\] is \[\frac{6}{{16}} = \frac{3}{8}\]
The person loses Rs. \[3.50\] when \[1\] head and \[3\] tails turn up, that is, when the events \[\left\{ {HTTT,THTT,TTHT,TTTH} \right\}\] occur.
Probability of losing Rs. \[3.50\] is \[\frac{4}{{16}} = \frac{1}{4}\] .
The person loses Rs. \[6\] when \[4\] tails turn up, that is, when the event \[\left\{ {TTTT} \right\}\] occurs.
Probability of losing Rs. \[6\] is \[\frac{1}{{16}}\] .
8. Three coins are tossed once. Find the probability of getting:
(i) \[\mathbf{3}\] heads
Ans. On tossing three coins, the sample space is obtained as:
\[S = \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \] .
Hence, \[n(S) = 8\]
Let B be the event that \[3\] heads occur.
So, \[B = \left\{ {HHH} \right\}\]
\[P(B) = \frac{{n(B)}}{{n(S)}}\]
\[ \Rightarrow P(B) = \frac{1}{8}\]
(ii) \[\mathbf{2}\] heads
Ans. Let C be the event that \[2\] heads occur.
So, \[C = \left\{ {HHT,{\rm{ }}HTH,THH} \right\}\]
\[P(C) = \frac{{n(C)}}{{n(S)}}\]
\[ \Rightarrow P(C) = \frac{3}{8}\]
(iii) At least \[\mathbf{2}\] heads
Ans. Let D be the event that at least \[2\] heads occur.
So, \[D = \left\{ {HHH,HHT,HTH,THH} \right\}\]
\[P(D) = \frac{{n(D)}}{{n(S)}}\]
\[ \Rightarrow P(D) = \frac{4}{8}\]
\[ \Rightarrow P(D) = \frac{1}{2}\]
(iv) At most \[\mathbf{2}\] heads
Ans. Let E be the event that at most \[2\] heads occur.
So, \[E = \left\{ {HHT,HTH,THH,HTT,THT,TTH,TTT} \right\}\]
\[P(E) = \frac{{n(E)}}{{n(S)}}\]
\[ \Rightarrow P(E) = \frac{7}{8}\]
(v) No heads
Ans. Let F be the event that no heads occur.
So, \[F = \left\{ {TTT} \right\}\]
\[P(F) = \frac{{n(F)}}{{n(S)}}\]
\[ \Rightarrow P(F) = \frac{1}{8}\]
(vi) \[\mathbf{3}\] tails
Ans. Let G be the event that \[3\] tails occur.
So, \[G = \left\{ {TTT} \right\}\]
\[P(G) = \frac{{n(G)}}{{n(S)}}\]
\[ \Rightarrow P(G) = \frac{1}{8}\]
(vii) Exactly \[\mathbf{2}\] tails
Ans. Let H be the event that exactly \[2\] tails occur.
So, \[H = \left\{ {HTT,THT,TTH} \right\}\]
\[P(H) = \frac{{n(H)}}{{n(S)}}\]
\[ \Rightarrow P(H) = \frac{3}{8}\]
(viii) No tail
Ans. Let I be the event that no tails occur.
So, \[I = \left\{ {HHH} \right\}\]
\[P(I) = \frac{{n(I)}}{{n(S)}}\]
\[ \Rightarrow P(I) = \frac{1}{8}\]
(ix) At most \[\mathbf{2}\] tails
Ans. Let J be the event that at most \[2\] tails occur.
So, \[J = \left\{ {HHT,HTH,THH,HTT,THT,TTH,TTT} \right\}\]
\[P(J) = \frac{{n(J)}}{{n(S)}}\]
\[ \Rightarrow P(J) = \frac{7}{8}\]
9. If \[\frac{2}{{11}}\] is the probability of an event, what is the probability of the event ‘not A’?
Ans. We are given that \[P(A) = \frac{2}{{11}}\] .
Consequently, \[P(A') = 1  P(A)\]
\[ \Rightarrow P(A') = 1  \frac{2}{{11}}\]
\[ \Rightarrow P(A') = \frac{9}{{11}}\]
10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that the letter is:
(i) A vowel
Ans. There are a total of \[13\] letters in the word ‘ASSASSINATION’. Hence \[n(S) = 13\] .
There are \[6\] vowels in the word give. Let A be the event that a vowel is chosen.
\[P(A) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow P(A) = \frac{6}{{13}}\]
(ii) A consonant
Ans. There are \[7\] consonants in the word given. Let B be the event that a vowel is chosen.
\[P(B) = \frac{{n(B)}}{{n(S)}}\]
\[ \Rightarrow P(B) = \frac{7}{{13}}\]
11. 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 lottery committee, he wins the price. What is the probability of winning the prize in the games?
(Hint: order of the numbers is not important.)
Ans. The total number of ways in which a person can choose six different numbers from \[1\] to \[20\] are:
\[^{20}{C_6} = \frac{{20!}}{{6!(20  6)!}}\]
\[{ \Rightarrow ^{20}}{C_6} = \frac{{20!}}{{6!14!}}\]
\[{ \Rightarrow ^{20}}{C_6} = \frac{{20 \times 19 \times 18 \times 17 \times 16 \times 15}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}\]
\[ \Rightarrow 38760\]
Hence, there are \[38760\] possible combinations of six numbers.
Out of all these combinations, one combination is fixed by the lottery committee.
Hence, the required probability of winning the lottery prize is \[\frac{1}{{38760}}\] .
12. Check whether the following probabilities P(A) and P(B) are consistently defined:
(i) \[\mathbf{P(A) = 0.5}\] , \[\mathbf{P(B) = 0.7}\] , \[\mathbf{P(A \cap B) = 0.6}\]
Ans. We know that \[P(A) = 0.5\] , \[P(B) = 0.7\] , \[P(A \cap B) = 0.6\] .
We know that if E and F are two events such that \[E \subset F\] , then \[P(E) \le P(F)\] .
In the given case, however, \[P(A \cap B) > P(A)\] . Hence, p(A) and P(B) are not consistently defined.
(ii) \[\mathbf{P(A) = 0.5}\] , \[\mathbf{P(B) = 0.4}\] , \[\mathbf{P(A \cap B) = 0.8}\]
Ans. We know that \[P(A) = 0.5\] , \[P(B) = 0.4\] , \[P(A \cap B) = 0.8\] .
We know that if E and F are two events such that \[E \subset F\] , then \[P(E) \le P(F)\] .
In the given case, \[P(A \cap B) > P(A)\] and \[P(A \cap B) > P(B)\] . Hence, \[P(A)\] and \[P(B)\] are consistently defined.
13. Fill in the blanks in the following table:
\[P(A)\]  \[P(B)\]  \[P(A \cap B)\]  \[P(A \cup B)\]  
(i)  \[\frac{1}{3}\]  \[\frac{1}{5}\]  \[\frac{1}{{15}}\]  … 
Ans. We know that \[P(A) = \frac{1}{3}\] , \[P(B) = \frac{1}{5}\] , \[P(A \cap B) = \frac{1}{{15}}\] .
\[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow P(A \cup B) = \frac{1}{3} + \frac{1}{5}  \frac{1}{{15}}\]
\[ \Rightarrow P(A \cup B) = \frac{7}{{15}}\]
\[P(A)\]  \[P(B)\]  \[P(A \cap B)\]  \[P(A \cup B)\]  \[P(A)\] 
(ii)  \[0.35\]  …  \[0.25\]  \[0.6\] 
Ans. We know that \[P(A) = 0.35\] , \[P(A \cup B) = 0.6\] , \[P(A \cap B) = 0.25\] .
\[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow 0.6 = 0.35 + P(B)  0.25\]
\[ \Rightarrow P(B) = 0.5\]
\[P(A)\]  \[P(B)\]  \[P(A \cap B)\]  \[P(A \cup B)\]  
(iii)  \[0.5\]  \[0.35\]  …  \[0.7\] 
Ans. We know that \[P(A) = 0.5\] , \[P(B) = 0.35\] , \[P(A \cup B) = 0.7\] .
\[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow 0.7 = 0.5 + 0.35  P(A \cap B)\]
\[ \Rightarrow P(A \cap B) = 0.15\]
14. Given \[P(A) = \frac{3}{5}\]and \[P(B) = \frac{1}{5}\] . Find P(A or B) , if A and B are mutually exclusive events.
Ans. We know that \[P(A) = \frac{3}{5}\] , \[P(B) = \frac{1}{5}\] .
For mutually exclusive events A and B,
\[P(A \cup B) = P(A) + P(B)\]
\[ \Rightarrow P(A \cup B) = \frac{3}{5} + \frac{1}{5}\]
\[ \Rightarrow P(A \cup B) = \frac{4}{5}\]
15. If E and F are events such that \[P(E) = \frac{1}{4}\] , \[P(F) = \frac{1}{2}\] and \[P(E \cap F) = \frac{1}{2}\] ,find:
(i) P(E or F)
Ans. Here, \[P(E) = \frac{1}{4}\] , \[P(F) = \frac{1}{2}\] , and \[P(E \cap F) = \frac{1}{8}\] .
We know that \[P(E \cup F) = P(E) + P(F)  P(E \cap F)\] .
So, \[P(E \cup F) = \frac{1}{4} + \frac{1}{2}  \frac{1}{8}\]
\[ \Rightarrow P(E \cup F) = \frac{5}{8}\]
(ii) P(not E and not F)
Ans. We know that \[P(E \cup F) = \frac{5}{8}\] .
Further, \[P(E \cup F)' = P(E' \cap F')\]
So, \[P(E' \cap F') = 1  P(E \cup F)\]
\[ \Rightarrow P(E' \cap F') = 1  \frac{5}{8}\]
\[ \Rightarrow P(E' \cap F') = \frac{3}{8}\]
16. Events E and F are such that P(Not E Not F) \[ = 0.25\] . State whether E and F are mutually exclusive.
Ans. We are given that P (not E or not F)=0.25 , that is, \[P(E' \cup F') = 0.25\]
Also, \[P(E' \cup F') = P(E \cap F)'\]
\[ \Rightarrow 0.25 = 1  P(E \cap F)\]
\[ \Rightarrow P(E \cap F) = 0.75\]
We note that \[P(E \cap F) \ne 0\] .
Therefore, E and F are not mutually exclusive.
17. A and B are events such that \[P\left( A \right) = 0.42\] , \[P\left( B \right) = 0.48\] and \[P\left( {A \cap B} \right) = 0.16\] . Determine:
(i) P(not A) \[ = 0.25\]
Ans. It is given that \[P(A) = 0.42\] , \[P(B) = 0.48\] and \[P(A \cap B) = 0.16\] .
\[P(A') = 1  P(A)\]
\[ \Rightarrow P(A') = 1  0.42\]
\[ \Rightarrow P(A') = 0.58\]
(ii) P(not B)
Ans. It is given that \[P(A) = 0.42\] , \[P(B) = 0.48\] and \[P(A \cap B) = 0.16\] .
\[P(B') = 1  P(B)\]
\[ \Rightarrow P(B') = 1  0.48\]
\[ \Rightarrow P(B') = 0.52\]
(iii) P(A or B)
Ans. It is given that \[P(A) = 0.42\] , \[P(B) = 0.48\] and \[P(A \cap B) = 0.16\] .
\[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow P(A \cup B) = 0.42 + 0.48  0.16\]
\[ \Rightarrow P(A \cup B) = 0.74\]
18. 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.
Ans. Let A be the event that the selected student studies Mathematics and B be the event that the selected student studies Biology.
Subsequently, \[P(A) = 40\% \]
\[ \Rightarrow \frac{{40}}{{100}} = \frac{2}{5}\]
Similarly, \[P(B) = 30\% \]
\[ \Rightarrow \frac{{30}}{{100}} = \frac{3}{{10}}\]
Further, \[P(A \cap B) = 10\% \]
\[ \Rightarrow \frac{{10}}{{100}} = \frac{1}{{10}}\]
We know that \[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow P(A \cup B) = \frac{2}{5} + \frac{3}{{10}}  \frac{1}{{10}}\]
\[ \Rightarrow P(A \cup B) = \frac{6}{{10}}\]
\[ \Rightarrow P(A \cup B) = 0.6\]
Thus, the probability that selected student studies Mathematics or Biology is \[0.6\] .
19. 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?
Ans. Let A and B be the events of passing the first and the second examinations respectively. Accordingly, \[P(A) = 0.8\] , \[P(B) = 0.7\] and \[P(A \cup B) = 0.95\]
We know that \[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow 0.95 = 0.8 + 0.7  P(A \cap B)\]
\[ \Rightarrow P(A \cap B) = 0.8 + 0.7  0.95\]
\[ \Rightarrow P(A \cap B) = 0.55\]
Thus, the probability of passing both these examinations is \[0.55\] .
20. 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?
Ans. Let A be the event of passing the English examination and B be the event of passing the Hindi examination.
Accordingly, \[P(A \cap B) = 0.5\] , \[P(A' \cap B') = 0.1\] and \[P(A) = 0.75\] .
Now, \[P(A' \cap B') = P(A \cup B)'\]
\[ \Rightarrow P(A \cup B)' = 0.1\]
Further, \[P(A \cup B)' = 1  P(A \cup B)\] .
\[ \Rightarrow 1  P(A \cup B) = 0.1\]
\[ \Rightarrow P(A \cup B) = 0.9\]
We know that \[P(A \cup B) = P(A) + P(B)  P(A \cap B)\] .
\[ \Rightarrow 0.9 = 0.75 + P(B)  0.5\]
\[ \Rightarrow P(B) = 0.9  0.75 + 0.5\]
\[ \Rightarrow P(B) = 0.65\]
Hence, probability of passing the Hindi examination is \[0.65\] .
21. 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.
Ans. Let A and B be the events that the selected student has opted for NCC and NSS respectively.
Total number of students \[ = 60\]
Number of students who have opted for NCC \[ = 30\]
\[P(A) = \frac{{n(A)}}{{n(S)}}\]
\[ \Rightarrow \frac{{30}}{{60}} = \frac{1}{2}\]
Number of students who have opted for NSS \[ = 32\]
\[P(B) = \frac{{n(B)}}{{n(S)}}\]
\[ \Rightarrow \frac{{32}}{{60}} = \frac{8}{{15}}\]
Number of students who have opted for both NCC and NSS \[ = 24\]
\[P(A \cap B) = \frac{{n(A \cap B)}}{{n(S)}}\]
\[ \Rightarrow \frac{{24}}{{60}} = \frac{2}{5}\]
We know that \[P(A \cup B) = P(A) + P(B)  P(A \cap B)\] .
\[ \Rightarrow P(A \cup B) = \frac{1}{2} + \frac{8}{{15}}  \frac{2}{5}\]
\[ \Rightarrow P(A \cup B) = \frac{{19}}{{30}}\]
Thus, the probability that the selected student has opted for NCC or NSS is \[\frac{{19}}{{30}}\] .
(ii) The student has opted neither NCC nor NSS.
Ans. We know that \[P(A' \cap B') = P(A \cup B)'\] .
\[ \Rightarrow P(A \cup B)' = 1  P(A \cup B)\]
\[ \Rightarrow 1  P(A \cup B) = 1  \frac{{19}}{{30}}\]
\[ \Rightarrow P(A \cup B)' = \frac{{11}}{{30}}\]
Thus, the probability that the selected student has opted for neither NCC nor NSS is \[\frac{{11}}{{30}}\] .
(iii) The student has opted NSS but not NCC.
Ans. We know that \[n(B  A) = n(B)  n(A \cap B)\] .
\[ \Rightarrow n(B  A) = 32  24\]
\[ \Rightarrow n(B  A) = 8\]
The number of students that have opted for NSS but not NCC is \[8\] .
\[P(B  A) = \frac{{n(B  A)}}{{n(S)}}\]
\[ \Rightarrow P(B  A) = \frac{8}{{60}}\]
\[ \Rightarrow P(B  A) = \frac{2}{{15}}\]
Thus, the probability that the selected student has opted for NSS but not for NCC is \[\frac{2}{{15}}\] .
Miscellaneous Exercise
1. 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?
Ans. Total number of marbles \[ = 60\]
The number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .
All the marbles drawn will be blue if we draw \[5\] marbles out of the \[20\] blue marbles. \[5\] blue marbles can be drawn out of the \[20\] blue marbles in \[^{20}{C_5}\] ways.
Probability that all marbles will be blue is \[\frac{{^{20}{C_5}}}{{^{60}{C_5}}}\] .
(ii) at least one will be green?
Ans. Total number of marbles \[ = 60\]
Number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .
Number of ways in which the drawn marbles is not green, that is, when red or blue marbles are drawn, is \[^{20 + 10}{C_5}{ = ^{30}}{C_5}\] .
So, probability that no marble is green is \[\frac{{^{30}{C_5}}}{{^{60}{C_5}}}\] .
Therefore, probability that at least one marble will be green is \[1  \frac{{^{30}{C_5}}}{{^{60}{C_5}}}\] .
2. \[4\] cards are drawn from a wellshuffled deck of \[52\] cards. What is the probability of obtaining three diamonds and one spade?
Ans. Number of ways of drawing \[4\] cards from a pack of \[52\] cards are \[^{52}{C_4}\] .
In a deck, there are \[13\] diamonds and \[13\] spades.
So, the number of ways of drawing three diamonds and one spade is \[^{13}{C_3}{ \times ^{13}}{C_1}\] .
Hence, probability of drawing three diamonds and one spade is \[\frac{{^{13}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_4}}}\] .
3. 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) \[P(2)\]
Ans. Total number of faces of the die is \[6\] .
Number of faces with the number ‘ \[2\] ’ is \[3\] .
Therefore, probability of obtaining the number \[2\] is \[P(2) = \frac{{n(2)}}{{n(S)}}\] .
\[ \Rightarrow P(2) = \frac{3}{6}\]
\[ \Rightarrow P(2) = \frac{1}{2}\]
(ii) \[\mathbf{P(1 \cup 3)}\]
Ans. Total number of faces of the die is \[6\] .
Number of faces with the number ‘ \[1\] ’ is \[2\] and with the number ‘ \[3\] ’ is \[1\] .
Therefore, probability of obtaining the number \[2\] or \[3\] is \[P(1 \cup 3) = \frac{{n(1 \cup 3)}}{{n(S)}}\] .
\[ \Rightarrow P(1 \cup 3) = \frac{{1 + 2}}{6}\]
\[ \Rightarrow P(1 \cup 3) = \frac{1}{2}\]
(iii) \[P(3')\]
Ans. Total number of faces of the die is \[6\] .
Number of faces with the number ‘ \[3\] ’ is \[1\] .
Therefore, probability of obtaining the number \[3\] is \[P(3) = \frac{1}{6}\] .
\[ \Rightarrow P(3') = 1  P(3)\]
\[ \Rightarrow P(3') = 1  \frac{1}{6}\]
\[ \Rightarrow P(3') = \frac{5}{6}\]
4. 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?
Ans. Total number of tickets sold is \[10,000\] .
Number of prizes awarded is \[10\] .
If A is the event we buy \[1\] ticket, the probability of winning will be \[P(A) = \frac{{n(A)}}{{n(S)}}\] .
\[ \Rightarrow P(A) = \frac{{10}}{{10,000}}\]
\[ \Rightarrow P(A) = \frac{1}{{1,000}}\]
So, probability of not getting a prize is \[P(A') = 1  P(A)\] .
\[ \Rightarrow P(A') = 1  \frac{1}{{1,000}}\]
\[ \Rightarrow P(A') = \frac{{999}}{{1,000}}\]
(ii) two tickets?
Ans. Total number of tickets sold is \[10,000\] .
Number of prizes awarded is \[10\] .
Number of tickets on which a prize is not awarded is \[10,000  10 = 9,990\] .
If B is the event we buy \[2\] tickets, the probability of not winning will be \[P(B) = \frac{{^{9,990}{C_2}}}{{^{10,000}{C_2}}}\] .
(iii) ten tickets?
Ans. Total number of tickets sold is \[10,000\] .
Number of prizes awarded is \[10\] .
Number of tickets on which a prize is not awarded is \[10,000  10 = 9,990\] .
If C is the event we buy \[10\] tickets, the probability of not winning will be \[P(C) = \frac{{^{9,990}{C_{10}}}}{{^{10,000}{C_{10}}}}\] .
5. 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
(i) You both enter the same section?
Ans. You and your friend are among the \[100\] students.
Total number of ways of selecting \[2\] students out of \[100\] students is ¹⁰⁰C₂.
If both of you enter the same section, you are either among the section having \[40\] students or the one having \[60\] students.
Number of ways in which you both can be in the same section is \[^{40}{C_2}{ + ^{60}}{C_2}\] .
Probability that you both can be in the same section is \[\frac{{^{40}{C_2}{ + ^{60}}{C_2}}}{{^{100}{C_2}}}\]
\[ \Rightarrow \frac{{\frac{{40!}}{{2!38!}} + \frac{{60!}}{{2!58!}}}}{{\frac{{100!}}{{2!98!}}}}\]
\[ \Rightarrow \frac{{(40 \times 39) + (60 \times 59)}}{{(100 \times 99)}}\]
\[ \Rightarrow \frac{{17}}{{33}}\]
(ii) You both enter different sections?
Ans. Probability that you both can be in the same section is \[\frac{{17}}{{33}}\] .
Probability that you both are not in the same section will be \[1  \frac{{17}}{{33}}\] .
\[ \Rightarrow \frac{{16}}{{33}}\]
6. 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 envelop contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Ans. Let \[{L_1},{L_2},{L_3}\] be the three letters and \[{E_1},{E_2},{E_3}\] be their corresponding envelopes.
There are \[6\] ways of inserting \[3\] letters in \[3\] envelopes. These are as follows:
\[\begin{array}{l}{L_1}{E_1},{L_2}{E_3},{L_3}{E_2}\\{L_2}{E_2},{L_1}{E_3},{L_3}{E_1}\\{L_3}{E_3},{L_1}{E_2},{L_2}{E_1}\\{L_1}{E_1},{L_2}{E_2},{L_3}{E_3}\\\;{L_1}{E_2},{L_2}{E_3},{L_3}{E_1}\\{L_1}{E_3},{L_2}{E_1},{L_3}{E_2}\end{array}\]
We observe that there are \[4\] ways in which at least one letter is inserted in its proper envelope.
Hence, the required probability is \[\frac{4}{6} = \frac{1}{2}\] .
7. A and B are two events such that \[P(A) = 0.54\] , \[P\left( B \right) = 0.69\] and \[P\left( {A \cap B} \right) = 0.35\] . Find:
(i) \[P\left( {A \cup B} \right)\]
Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .
We know that \[P(A \cup B) = P(A) + P(B)  P(A \cap B)\]
\[ \Rightarrow P(A \cup B) = 0.54 + 0.69  0.35\]
\[ \Rightarrow P(A \cup B) = 0.88\]
(ii) \[P\left( {A' \cap B'} \right)\]
Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .
We know that \[P\left( {A' \cap B'} \right) = P(A \cup B)'\]
\[ \Rightarrow P\left( {A' \cap B'} \right) = 1  P(A \cup B)\]
\[ \Rightarrow P\left( {A' \cap B'} \right) = 1  0.88\]
\[ \Rightarrow P\left( {A' \cap B'} \right) = 0.12\]
(iii) \[P\left( {A \cap B'} \right)\]
Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .
We know that \[P\left( {A \cap B'} \right) = P\left( A \right)P\left( {A \cap B} \right)\]
\[ \Rightarrow P\left( {A \cap B'} \right) = 0.540.35\]
\[ \Rightarrow P\left( {A \cap B'} \right) = 0.19\]
(iv) \[P\left( {B \cap A'} \right)\]
Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .
We know that \[P\left( {B \cap A'} \right) = P\left( B \right)P\left( {A \cap B} \right)\]
\[ \Rightarrow P\left( {B \cap A'} \right) = 0.690.35\]
\[ \Rightarrow P\left( {B \cap A'} \right) = 0.34\]
8. 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?
Ans. Let E be the event that the spokesperson will be a male and F be the event that the spokesperson will be over \[35\] years of age.
Subsequently, \[P(E) = \frac{3}{5}\] and \[P(F) = \frac{2}{5}\] .
Since there is only one male who is above \[35\] years of age, \[P(E \cap F) = \frac{1}{5}\] .
We know that \[P(E \cup F) = P(E) + P(F)  P(E \cap F)\]
\[ \Rightarrow P(E \cup F) = \frac{3}{5} + \frac{2}{5}  \frac{1}{5}\]
\[ \Rightarrow P(E \cup F) = \frac{4}{5}\]
Hence, the probability that the spokesperson will either be a male or over \[35\] years of age is \[\frac{4}{5}\] .
9. If four digits 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?
Ans. Since fourdigit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . Remaining three digits can be filled by any of the given digits with repetition.
So, total number of ways of forming fourdigit numbers are \[2 \times 5 \times 5 \times 5\] , and subtract \[1\] from the obtained answer, as we cannot count \[5,000\] in the numbers.
\[ \Rightarrow 250  1\]
\[ \Rightarrow 249\] ways
A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\] .
Total number of fourdigit numbers greater than \[5,000\] that are divisible by \[5\] are \[(2 \times 5 \times 5 \times 2)  1\] .
\[ \Rightarrow 100  1\]
\[ \Rightarrow 99\]
Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{99}}{{249}} = \frac{{33}}{{83}}\]
(ii) the repetition of digits is not allowed?
Ans. Since fourdigit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . Remaining three digits can be filled by any of the given digits without repetition.
So, total number of ways of forming fourdigit numbers are \[2 \times 4 \times 3 \times 2 = 48\] .
A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\] .
When the digit at the thousands place is \[5\] , the units place can be filled only with \[0\] , so the number of possible ways are \[1 \times 3 \times 2 \times 1 = 6\] .
When the digit at the thousands place is \[7\] , the units place can be filled only with \[0\] or \[5\] , so the number of possible ways are \[1 \times 3 \times 2 \times 2 = 12\] .
Total number of fourdigit numbers greater than \[5,000\] that are divisible by \[5\] are \[12 + 6 = 18\] .
Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{18}}{{48}} = \frac{3}{8}\] .
10. 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?
Ans. The number lock of suitcase has \[4\] wheels, each of which is labelled with ten digits, that is, from \[0\] to \[9\] .
Number of ways of selecting \[4\] different digits out of \[10\] digits is \[^{10}{C_4}\] .
Further, each combination of \[4\] different digits can be arranged in \[4\] ways.
Number of four digits with no repetitions \[{ = ^{10}}{C_4} \times 4!\]
\[ \Rightarrow \frac{{10!}}{{4!6!}} \times 4!\]
\[ \Rightarrow 10 \times 9 \times 8 \times 7\]
\[ \Rightarrow 5040\]
There is only one number that can open the suitcase.
Thus, the required probability is \[\frac{1}{{5040}}\] .
NCERT Solutions for Class 11 Maths Chapters
NCERT Solutions for Class 11 Maths Chapter 14 – All Exercises
Often, we use Probability assessments in our daily lives without even thinking too much about it. Likewise, researchers take their time finding out the precise Probability of helping health care providers, insurance companies, stakeholders, and many more critical decisions when it comes to using an informal Probability manner.
Probability can be different depending on how you try to get to the answer. If you are considering outcomes over a long span of repetition, it will come out to be 50:50. This illustrates an important point about how Probability works.
With our Probability Class 11 NCERT solutions PDF download, you can quickly get the answers to all the questions that you have in the textbook. All these questions are solved by experts in mathematics and have years of teaching experience. As a result, you will be getting all the little details and tricks to quickly solve these questions.
NCERT Solution Class 11 Maths of Chapter 14 All Exercises
Chapter 14  Probability Exercises in PDF Format  
16 Questions & Solutions  
7 Questions & Solutions  
21 Questions & Solutions  
Miscellaneous Exercise  10 Questions & Solutions 
NCERT Solutions of Class 11 Maths Chapter 14
The Class 11 Maths Probability solutions are easy to download, and once you download them, you can access the answers whenever you want without needing the internet. These NCERT solutions of Probability Class 11 will help you get good grades in your Maths examination.
14: Probability
14.1: Introduction
14.2: Random Experiments
14.3: Event
14.4: Axiomatic Approach to Probability
NCERT Solutions For Class 11 Chapter 14 Probability Weightage Marks
This chapter can help you score good marks in Maths exams. You can expect 12 to 10 marks questions to come in the exam from this chapter. Even in JEE mains, you can get 23 questions from Probability, so you need to take this chapter seriously when you are preparing for the competitive exam.
Topics covered in Class 11 Maths Chapter 14 Probability:
14.1 Introduction
Probability is a measure to tell how likely an event is going to happen. In mathematical terms, it is defined as the ratio of the number of favorable outcomes of an event to the total number of outcomes of the event. For an experiment having a total 'n' number of possible outcomes, the ‘x’ number of favorable outcomes, then the formula to calculate the probability of an event is as follows.
Probability of an event = Favorable Outcomes/Total Outcomes = x/n
The number of favorable outcomes can never cross the total number of outcomes.
The probability value of any outcome in an experiment will always lie between 0 and 1. Where the probability 0 represents that the event is not going to happen and the probability 1 represents the event happens every time.
The value of probability of any outcome in an experiment can never be negative.
14.2 Random Experiments
In Probability, Random experiments are defined as experiments that may be repeated several times under the same conditions and the outcome of which cannot be predicted with 100% certainty.
14.3 Event
An Event in probability is a set of outcomes of any random experiment. The sample space includes all possible outcomes of an experiment. So, we can observe that the events are the subsets of the sample space.
14.4 Axiomatic Approach to Probability
Axiomatic Probability is another approach of describing the probability of an event. In this type of probability, few axioms (rules) are predefined before assigning probabilities. In this method, we can ease the calculation of occurrence and nonoccurrence of the event.
Benefits of Probability Class 11 NCERT Solutions
With this study material, you will be able to learn the chapter well and make better decisions.
Class 11 Probability NCERT solutions will make the student's decision capability strong.
Another time you will be using Probability is in your car insurance. Companies put the base of their insurance premiums on the Probability of your car meeting an accident.
Moreover, in the ch 14 Maths Class 11, you will know about the Probability theory, which plays an important role when making predictions.
When it comes to research investigation, predictions and estimates play a vital role in the inquiring lead to correct decision making.
The Class 11 Probability will be used in several ways. One widespread use is the Probability of developing cancer.
In addition to this, when it comes to getting good marks in your Maths exam, this chapter can help you reach your goal without solving complex questions that take a lot of time. A question from the Probability chapter is like free marks for students in a question paper.
So, with the help of Probability Class 11 PDF download, you will know how to solve Probability problems during exams.
FAQs on NCERT Solutions for Class 11 Maths Chapter 14  Probability
1. What is a sample space and how does it help?
A set consists of all the possible outcomes of an upcoming event, such as taking a card out from the deck of cards. It can be defined as how you can choose something from the given sample and sample space. Every single Probability is a small portion of the sample space. As a result, the sum of all the possibilities will come out to be one. In addition to this, if you are trying to find an event's occurrence, you will see that it is always one minus the Probability that didn't occur.
2. What do you mean by Conditional Probability?
A Probability condition in which the event will only occur after confirmation that the first event has already happened. In these types of cases, both the events that will occur are independent of each other. As a result, event A will not affect event B's probability, making Event A simply the Probability of event B.
3. Where can I download the latest NCERT Solutions for Class 11 Chapter 14 Probability?
The latest NCERT book for Class 11 Probability can be found on the NCERT site. The solutions for the same are NCERT solutions Class 11 Chapter 14 Probability. It is advisable that prior to referring to the solutions, the student tries solving the exercises on their own rather than just looking up the answers. Any further help can be found on the Vedantu site for achieving better marks in the exams.
4. Can I download the NCERT Solutions for Class 11 Probability in PDF?
The solutions are available free of cost on the Vedantu site.
Visit the pageNCERT solutions Class 11 and choose Chapter 14 Probability.
The webpage with Vedantu’s solutions for the NCERT ebook for Class 11 Probability will open.
To download the PDF, click on the Download PDF button and you can view the solutions offline.
The Vedantu website has a variety of study resources that can aid you with this or any other chapter.
5. What is the significance of the Random Experiment?
A Random experiment is a test, trial, or observation that may be repeated many times under the same conditions. An individual random experiment's outcome must be independent and uniformly distributed. It is an important aspect of the Chapter of Probability in Class 11 Maths Chapter 14. Refer to modules provided on the site of Vedantu to get a better understanding of the various topics of Probability.
6. Are the exercises given in the Probability Chapter important?
Yes, it is necessary to do so. These questions will aid in bridging your knowledge gap and give you an idea as to what the concepts are in which you need to work harder. Moreover, the final paper and the way it is set will have a resemblance to these exercises and hence you would be familiar with your exam as well. Modules relating to this topic or other topics covered in this chapter may be found on the Vedantu website or the Vedantu app.
7. How many Chapters are there in the NCERT Solutions of Class 11 Maths apart from Chapter 14?
There are a total of 14 chapters, each with its own distinct and distinctive material, and it is recommended that the student thoroughly prepare each chapter and comprehend all of the themes in order to perform well in his tests. If you need assistance, you may go to the Vedantu app for NCERT Class 11 Maths Chapter 14 and other chapter’s answers. Modules on this or other subjects in this chapter can be found on the Vedantu website or the Vedantu mobile application.