NCERT Solutions  Class 11 Maths Chapter 16  Probability
CBSE Solutions  Maths Class 11
Check the below NCERT Solutions for Class 11 Maths Chapter 16 Probability Pdf free download. NCERT Solutions Class 11 Maths were prepared based on the latest exam pattern. We have Provided Probability Class 11 Maths NCERT Solutions to help students understand the concept very well.
NCERT  Class 11 Maths
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Maths 
Chapter:  16 
Chapters Name:  Probability 
Medium:  English 
Probability  Class 11 Maths  NCERT Books Solutions
NCERT Solutions for Class 11 Maths Chapter 16 Probability Ex 16.1
In each of the following Exercises 1 to 7, describe the sample space for the indicated experiment.
Ex 16.1 Class 11 Maths Question 1.
A coin is tossed three times.
Solution.
When one coin is tossed three times, The sample space of the experiment is given by S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}.
Ex 16.1 Class 11 Maths Question 2.
A die is thrown two times.
Solution.
When a die is thrown two times. The sample space S for this experiment is given by
S = {(1, 1), (1, 2),(1, 6), (2, 1), (2, 2), … (2, 6), …, (6, 1),…, (6, 6)}.
Ex 16.1 Class 11 Maths Question 3.
A coin is tossed four times.
Solution.
When a coin is tossed four times. The sample space S for this experiment is given by
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}.
Ex 16.1 Class 11 Maths Question 4.
A coin is tossed and a die is thrown.
Solution.
When a coin is tossed and a die is thrown. The sample space S for the experiment is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.
Ex 16.1 Class 11 Maths Question 5.
A coin is tossed and then a die is rolled only in case a head is shown on the coin.
Solution.
When a coin is tossed and a die is rolled only in case if head is shown on the coin. The sample space S for the experiment is given by
S = {H1, H2, H3, H4, H5, H6, T}.
Ex 16.1 Class 11 Maths Question 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.
Solution.
Let B_{1}, B_{2} and G_{1}, G_{2} are the boys and girls respectively in room X, B_{3} and G_{3}, G_{4}, G_{5} are the boy and girls respectively in room Y. The sample space S for the experiment is given by
S = {XB_{1}, XB_{2}, XG_{1}, XG_{2}, YB_{3}, YG_{3}, YG_{4}, YG_{5}).
Ex 16.1 Class 11 Maths Question 7.
One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.
Solution.
Let R, W and B denote the red, white and blue dice respectively. The sample space S, for this experiment is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}.
Ex 16.1 Class 11 Maths Question 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?
(ii) What is the sample space if we are interested in the number of girls in the family?
Solution.
(i) The sample space S, in knowing whether it is a boy or a girl in the order of their births in composition of families with two children is S = {BB, BG, GB, GG}.
(ii) The sample space S, in knowing the number of girls in a family is S = {0,1, 2}.
Ex 16.1 Class 11 Maths Question 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.
Solution.
There are 1 red and 3 identical white balls in a box. The sample space S for this experiment is given by S = {RW, WR, WW}.
Ex 16.1 Class 11 Maths Question 10.
An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.
Solution.
The sample space S, for tossing a coin and then tossing it second time if a head occurs; if a tail occurs on the first toss, the die is tossed once is given by S = {HH, HT, T1, T2, T3, T4, T5, T6}.
Ex 16.1 Class 11 Maths Question 11.
Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or nondefective (N). Write is the sample space of this experiment?
Solution.
The sample space S for selecting three bulbs at random from a lot is
S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}.
Ex 16.1 Class 11 Maths Question 12.
A coin is tossed. If the out come is a 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?
Solution.
An experiment consists of tossing a coin. If the result is a head, a die is thrown. If the die shows up an even number, the die is thrown again. The sample space S for this experiment is
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}.
Ex 16.1 Class 11 Maths Question 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.
Solution.
Four slips marked as 1,2,3 and 4 are put in a box. Two slips are drawn from it one after the other without replacement. The sample space S, for the experiment is S = {(1, 2), (1, 3), (1, 4), (2,1), (2, 3), (2, 4), (3,1), (3, 2), (3, 4), (4,1), (4, 2), (4, 3)}.
Ex 16.1 Class 11 Maths Question 14.
An experiment consists of rolling a die and then tossing a coin once 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.
Solution.
An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number is odd, the coin is tossed twice. The sample space S for this experiment is given by S = {1HH, 1TH, 1HT, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}.
Ex 16.1 Class 11 Maths Question 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.
Solution.
An experiment consists of tossing a coin. If it shows a tail, a ball is drawn from a box which contains 2 red and 3 black balls. If it shows head, a die is thrown. Then the sample S for this experiment is given by
S = {TR_{1}, TR_{2}, TB_{1}, TB_{2}, TB_{3}, H_{1}, H_{2}, H_{3}, H_{4}, H_{5}, H_{6})
Ex 16.1 Class 11 Maths Question 16.
A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?
Solution.
An experiment consists of rolling a die.
∴ Sample space = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6),…, (1, 5, 6), (2, 1, 6), (2, 2, 6),…, (2, 5, 6),…, (5, 1, 6), (5, 2, 6),…}.
NCERT Solutions for Class 11 Maths Chapter 16 Probability Ex 16.2
Ex 16.2 Class 11 Maths Question 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?
Solution.
An experiment consists of rolling a die.
∴ S = {1, 2, 3, 4, 5, 6}
E: die shows 4 = {4}
F : die shows an even number = {2, 4, 6}
∴ E ∩F={4} ⇒ E∩F ≠ ⏀
⇒ E and F are not mutually exclusive.
Ex 16.2 Class 11 Maths Question 2.
A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) 8: 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 ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D E, E ∩ F’, F’.
Solution.
An experiment consists of rolling a die.
S = {1, 2, 3, 4, 5, 6}
(i) A: a number less than 7 = {1, 2, 3, 4, 5, 6}
(ii) B: a number less than 7 = ⌽
(iii) C: a multiple of 3 = {3, 6}
(iv) D : a number less than 4 = {1, 2, 3}
(v) E : an even number greater than 4 = {6}
(vi) F : a number not less than 3 = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6) ∩ ⌽
= {1, 2, 3, 4, 5, 6!
A ∩ B = {1, 2,3,4, 5, 6) ∩ ⌽ = ⌽
B ∪C = ⌽∪{3,6} = {3,6}
E ∩ F = {6} ∩ {3, 4, 5, 6) = {6}
D ∩ E = {1,2, 3} ∩ (6} = ⌽
A – C = (1, 2, 3, 4, 5, 6) – {3, 6} = {1, 2, 4, 5}
D – E = {1, 2, 3} – {6} = {1, 2, 3}
F’ = {1, 2, 3, 4, 5, 6) – {3, 4, 5, 6) = {1, 2)
E ∩F’=(6)∩{l, 2}= ⌽
Ex 16.2 Class 11 Maths Question 3.
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?
Solution.
An experiment consists of rolling a pair of dice.
∴ Sample space consists 6 x 6 = 62 = 36 possible outcomes.
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)}
Now, A : the sum is greater than 8
= {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B : 2 occurs on either die = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}
C : The sum is at least 7 and a multiple of 3 = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
A∩B =⌽, B∩C = ⌽
Thus above shows that A and B; B and C are mutually exclusive events.
Ex 16.2 Class 11 Maths Question 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?
(ii) Simple?
(iii) Compound?
Solution.
An experiment consists of tossing threecoins:
∴ S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ A : Three heads show = {HHH}
B : Two heads and one tail show = {HHT, HTH, THH}
C : Three tail show = {TTT}
D : A head show on the first coin = {HHH, HHT, HTH, HTT}
(i) Since A∩B = ⌽, A∩C = ⌽, B ∩ C = ⌽,
C ∩ D = ⌽.
⇒ A and B; A and C; B and C; C and D are mutually exclusive events.
(ii) A and C are simple events.
(iii) B and D are compound events.
Ex 16.2 Class 11 Maths Question 5.
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.
Solution.
An experiment consists of tossing three coins then the sample space S is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two events A and B which are mutually exclusive are
A : “getting atmost one head” and B: “getting atmost one tail”
(ii) Three events A, B and C which are mutually exclusive and exhaustive are
A : “getting atleast two heads”
B : “getting exact two tails” and C: “getting exactly three tails”
(iii) Two events A and B which are not mutually exclusive are
A : “getting exactly two tails” and B: “getting atmost two heads”
(iv) Two events A and B which are mutually exclusive but not exhaustive are
A : “getting atleast two heads” and B: “getting atleast three tails”
(v) Three events A, B and C which are mutually exclusive but not exhaustive are
A : “getting atleast three tails”
B : “getting atleast three heads”
C : “getting exactly two tails”
Ex 16.2 Class 11 Maths Question 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 ≤ 5.
Describe the events
(i) A’
(ii) not B
(iii) A or B
(iv) A and B
(v) A but bot C
(vi) B or C
(vii) B and C
(viii) A ∩ B’ ∩C’
Solution.
An experiment consists of rolling two dice Sample space consists 6 x 6 = 36 outcomes.
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (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)}
A : getting an even number on the first die = {(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 : getting an odd number on the first die = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, b), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, b), (5,1), (5, 2), (5, 3), (5,4), (5, 5), (5,6))
C: getting the sum of the numbers on the dice ≤5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (4, 1)}
(i) A’: getting an odd number on the first die=B
(ii) not B : getting an even number on the first die = A
(iii) A or B = A∪B = S
∴ A ∪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), (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)}}
(iv) A and B = A ∩ B = ⌽
(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 = BuC = {(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), B or C = BuC = {(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∩C = {(1,1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
(viii) A : getting an even number on the first die = B’
B’: getting an even number on the first die = {(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)}
C : getting the sum of numbers on two dice > 5. {(l, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (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 ∩ B’∩ 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)}
Ex 16.2 Class 11 Maths Question 7.
Refer to question 6 above, state true or false : (give reason for your answer).
(i) A and B are mutually exclusive
(ii) A and 6 are mutually exclusive and exhaustive
(iii) A = B’
(iv) A and C are mutually exclusive
(v) A and S’ are mutually exclusive.
(vi) A’, B’, C are mutually exclusive and exhaustive.
Solution.
(i) True.
A = getting an even number on the first die.
B = getting an odd number on the first die. There is no common elements in A and B.
⇒ A ∩ B = ⌽
∴ A and B are mutually exclusive.
(ii) True.
From (i), A and B are mutually exclusive.
A ∪ B = {(1, 1), (1, 2) (1, 6), (2,1), (2, 2), (2, 6),…, (6,1), (6, 2), …, (6, 6) = S
∴ A∪B is mutually exhaustive.
(iii) True.
B = getting an odd number on the first die.
B’ = getting an even number on first die = A.
∴ A = B’
(iv) False.
Since A ∩ C={(2, 1), (2, 2), (2, 3), (4, 1)}
(v) False.
Since B’ = A [from (iii)]
∴ A∩B’=A∩A = A ≠ ⌽
(vi) False.
Since A’ = B and B’=A, A’ ∩ B’ = ⌽
B ∩ C = {(1,1), (1,2), (1, 3), (1,4), (3,1), (3, 2)} ≠ ⌽
A ∩ C = {(2, 1), (2, 2), (2, 3), (4,1)} ≠ ⌽
Thus A’, B’ and C are not mutually exclusive.
NCERT Solutions for Class 11 Maths Chapter 16 Probability Ex 16.3
Ex 16.3 Class 11 Maths Question 1.
Which of the following cannot be valid assignment of probabilities for outcomes of sample space
S = {⍵_{1}, ⍵_{2}, ⍵_{3}, ⍵_{4}, ⍵_{5}, ⍵_{6}, ⍵_{7}}
Solution.
(a) Sum of probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00
∴ Assignment of probabilities is valid.
(b) Sum of probabilities
\(=\frac { 1 }{ 7 } +\frac { 1 }{ 7 } +\frac { 1 }{ 7 } +\frac { 1 }{ 7 } +\frac { 1 }{ 7 } +\frac { 1 }{ 7 } +\frac { 1 }{ 7 } =\frac { 7 }{ 7 } =1\)
∴ Assignment of probabilities is valid.
(c) Sum of probabilities
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8
Sum of probabilities is greater than 1.
∴ This assignment of probabilities is not valid.
(d) Probability of any event cannot be negative. Therefore, this assignment of probabilities is not valid.
(e) The last probability \(\frac { 15 }{ 14 } \) is greater than 1.
∴ This assignment of probabilities is not valid.
Ex 16.3 Class 11 Maths Question 2.
A coin is tossed twice, what is the probability that atleast one tail occurs?
Solution.
An experiment consists of tossing a coin twice.
The sample space of the given experiment is given by
S = {HH, HT, TH, TT}
Let E be the event of getting atleast one tail.
Then, E = {HT, TH, TT}
∴ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 3 }{ 4 } \)
 Probability Class 11 Ex 16.1
 Probability Class 11 Ex 16.2
Ex 16.3 Class 11 Maths Question 3.
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.
Solution.
An experiment consists of throwing a die.
∴ The sample space of the experiment is given by S = {1, 2, 3, 4, 5, 6}
(i) Let E be the event that a prime number will appear.
∴ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let F be the event that a number ≥ 3 will appear.
F ={3, 4, 5, 6}
∴ \(P\left( E \right) =\frac { n\left( F \right) }{ n\left( S \right) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
(iii) Let G be the event that a number ≤ 1 will appear.
G={l}.
∴ \(P\left( G \right) =\frac { n\left( G \right) }{ n\left( S \right) } =\frac { 1 }{ 6 } \)
(iv) Let H be the event that a number more than 6 will appear.
H = ⌽
∴ \(P\left( H \right) =\frac { n\left( H \right) }{ n\left( S \right) } =\frac { 0 }{ 6 } =0\)
(v) Let I be the event that a number less than 6 will appear.
I = (1, 2, 3, 4, 5}
∴ \(P\left( I \right) =\frac { n\left( I \right) }{ n\left( S \right) } =\frac { 5 }{ 6 } \)
Ex 16.3 Class 11 Maths Question 4.
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.
Solution.
(a) There are 52 cards in a pack.
⇒ Number of points in the sample space S = n(S) = 52
(b) Let E be the event of drawing an ace of spades.
There is only one ace of spade n(E) = 1 and n(S) = 52
∴ \([P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 1 }{ 52 } \)
(c)
(i) Let F be the event of drawing an ace. There are 4 aces in a pack of 52 cards. n(F) = 4, n(S) = 52
∴ \(P\left( F \right) =\frac { n\left( F \right) }{ n\left( s \right) } =\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
(ii) Let G be the event of drawing a black card. There are 26 black cards. n(G) = 26, n(S) = 52
∴ \(P\left( G \right) =\frac { n\left( G \right) }{ n\left( S \right) } =\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
Ex 16.3 Class 11 Maths Question 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
(ii) 12.
Solution.
An experiment consists of tossing a coin marked 1 and 6 on either faces and rolling a die.
∴ The sample space of the experiment 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)}
(i) Let E be the event that sum of number is 3.
E = {(1,2)} ⇒ n(E) = 1
n(S)= 12
∴ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 1 }{ 12 } \)
(ii) Let F be the event that sum of number is 12.
∴ F = {(6, 6)} ⇒ n(F) = 1 and n(S) = 12
⇒ \(P\left( E \right) =\frac { n\left( F \right) }{ n\left( S \right) } =\frac { 1 }{ 12 } \)
Ex 16.3 Class 11 Maths Question 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?
Solution.
There are 6 women and 4 men.
An experiment consists of selecting a council member at random.
∴ n(S) = 10
Let E be the event that the selected council member will be a woman.
n(E) = 6
∴ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 6 }{ 10 } =\frac { 3 }{ 5 } \)
Ex 16.3 Class 11 Maths Question 7.
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.
Solution.
An experiment consists of tossing a fair coin four times. Therefore, the sample space of the given experiment is given by S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THTH, TTHH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
According to question, we have
Ex 16.3 Class 11 Maths Question 8.
Three coins are tossed once.Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) atleast 2 heads
(iv) atmost 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) atmost two tails
Solution.
An experiment consists of tossing 3 coins
∴ The sample space of the given experiment is given by
S = {HHH, HHT, HTH, THH, TTH, THT, HU ITT}
∴ n(S) = 8
Ex 16.3 Class 11 Maths Question 9.
If \(\frac { 2 }{ 11 } \) is the probability of an event, what is the probability of the event’not A’.
Solution.
Let P(A) = \(\frac { 2 }{ 11 } \)
P(not A) = 1 – P(A) = \(1\frac { 2 }{ 11 } =\frac { 9 }{ 11 } \).
Ex 16.3 Class 11 Maths Question 10.
A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel
(ii) a consonant.
Solution.
An experiment consists of a letter chosen at random from the word ‘ASSASSINATION’ which consists 13 letters,
(6 vowels and 7 consonants).
∴ Sample points are 13.
(i) Let E be the event that chosen letter is a vowel
E = {A, A, A, I, I, O}
∴ n(E) = 6
⇒ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 6 }{ 13 } \).
(ii) Let E be the event that chosen letter is a consonant
∴ F = {S, S, S, S, N, N, T}
⇒ \(P\left( F \right) =\frac { n\left( F \right) }{ n\left( S \right) } =\frac { 7 }{ 13 } \)
Ex 16.3 Class 11 Maths Question 11.
In a lottery, a person chosen 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]
Solution.
An experiment consists of a lottery, a person chose six different natural numbers at random from 1 to 20.
∴ Sample points
= ^{26}C6 = \(\frac { 20\times 19\times 18\times 17\times 16\times 15 }{ 1\times 2\times 3\times 4\times 5\times 6 } =38760\)
Let E be the event that chosen six numbers match with the six numbers already fixed by the lottery committee, i.e. Winning the prize, in the game
n(E) = ^{6}C_{6} = 1
∴ \(P\left( E \right) =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 1 }{ 38760 } \)
Ex 16.3 Class 11 Maths Question 12.
Check whether the following probabilities P(A) and P(B) are consistently defined.
(i) P(4) = 0.5, P(B) = 0.7, P(A∩B) = 0.6
(ii) P(A) = 0.5, P(S) = 0.4, P(A ∪ B) = 0.8
Solution.
(i) P(A ∩ B) must be less than or equal to P(A) and P(B)
∴ P(A ∩ B) = 0.6 > 0.5 = P(A)
∴ P(A) and P(B) are not defined consistently.
(ii) P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 0.5 + 0.4 – 0.8
= 0.9 – 0.8 = 01
∴ P(A ∩B) = 0.1 < 0.5 = P(A)
and P(A ∩ B) = 0.1 < 0.4 = P(B)
Thus, P(A) and P(B) are consistently defined.
Ex 16.3 Class 11 Maths Question 13.
Fill in the blanks in following table:
Solution.
(i) P(A ∪B) = P(A) + P(B) – P(A∩B)
\(\frac { 1 }{ 3 } +\frac { 1 }{ 5 } \frac { 1 }{ 15 } =\frac { 5+31 }{ 15 } =\frac { 7 }{ 15 } \)
(ii) P(A∪B) = P(A) + P(B) – p(A ∩B)
⇒ 0.6 = 0.35 + P(B) – 0.25
∴P(B) = 0.6 – 0.35 + 0.25 = 0.5
(iii) P(A∪B) = P(A) +P(B) – P(A∩B)
⇒ 0.7 = 0.5 + 0.35 – P(A∩B)
∴P(A∩B) = 0.5 + 0.35 – 0.7 = 0.15
Ex 16.3 Class 11 Maths Question 14.
Given P(4) = \(\frac { 3 }{ 5 } \) and P(B) = \(\frac { 1 }{ 5 } \) Find P{A or B), if A and B are mutually exclusive events.
Solution.
When A and B are mutually exclusive events.
⇒ A ∩ B = ⌽
⇒ P(A ∩ B) = 0
∴ P(A∪B) = P(A) + P(B) = \(\frac { 3 }{ 5 } +\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Ex 16.3 Class 11 Maths Question 15.
If E and Fare events such that P(E) = \(\frac { 1 }{ 4 } \), P(F) = \(\frac { 1 }{ 2 } \) and
P(E andF) = \(\frac { 1 }{ 8 } \), find
(i) P(E or F),
(ii) P(not E and not F).
Solution.
(i) P(E or F) = P(E ∪F)
= P(E) + P(F) – P(E ∩F)
\(=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } \frac { 1 }{ 8 } =\frac { 2+41 }{ 8 } =\frac { 5 }{ 8 } \)
(ii) not E and not F = E’ ∩ F’ = (E ∩ F)’
(De Morgan’s Law)
∴ P(not E and not F) = P(E ∪ F)’
=1 – P(E∪F) = \(1\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)
Ex 16.3 Class 11 Maths Question 16.
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
Solution.
not E or not F = E’ ∪ F’ = (E ∩ F)’
(De Morgan’s Law)
∴ P(not E or not F) = P(E ∩ F)’ = 1 – P (E ∩ F)
⇒ 0.25 = 1 – P(E ∩ F)
⇒ P(E ∩ F) = 1 – 0.25 = 0.75 ≠ 0
∴ Events E and F are not mutually exclusive.
Ex 16.3 Class 11 Maths Question 17.
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)
Solution.
(i) P(not A) = P(A) = 1 P(A) = 1 0.42 = 0.58
(ii) P(not B) = P(B’) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B)
= P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.74.
Ex 16.3 Class 11 Maths Question 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.
Solution.
Let E and F be the events that students study Mathematics and Biology respectively. Probability
that students study Mathematics i.e.,
\(P\left( E \right) =\frac { 40 }{ 100 } =0.4\)
Probability that students study Biology i.e.,
\(P\left( F \right) =\frac { 30 }{ 100 } =0.3\)
Probability that students study both Mathematics and Biology i.e.,
\(P\left( E\cap F \right) =\frac { 10 }{ 100 } =0.1\)
We have to find the probability that a student studies Mathematics or Biology, i.e., P(E ∪ F)
Now, P(E ∪ F) = 0.4 + 0.3 – 0.1 = 0.6
Ex 16.3 Class 11 Maths Question 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 atleast one of them is 0.95. What is the probability of passing both?
Solution.
Let E be the event that the student passes the first examination and F be the event that the student passes the second examination. Then P(E) = 0.8, P(F) = 0.7, and P(E u F) = 0.95 We know that
P(E ∪F) = P(E) + P(F) – P(E ∩ F)
⇒ 0.95 = 0.8 + 0.7 P(E∩F)
⇒ 0.95 = 1.5 – P(E∩F)
∴ P(E∩F) = 1.5 – 0.95 = 0.55.
Ex 16.3 Class 11 Maths Question 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?
Solution.
Let E be the event that student passes English examination and F be the event that the student passes Hindi examination.
Ex 16.3 Class 11 Maths Question 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.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Solution.
Here total number of students, n(S) = 60 Let E be the event that student opted for
NCC and F be the event that the student opted for NSS.
Then n(E) = 30, n(F) = 32 and n(E ∩ F) = 24
NCERT Class 11 Maths
Class 11 Maths Chapters  Maths Class 11 Chapter 16
Chapterwise NCERT Solutions for Class 11 Maths

NCERT Solutions For Class 11 Maths Chapter 1 Sets
NCERT Solutions For Class 11 Maths Chapter 2 Functions and relations
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric functions
NCERT Solutions For Class 11 Maths Chapter 4 Principle of mathematical induction
NCERT Solutions For Class 11 Maths Chapter 5 Quadratic equations and complex numbers
NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions For Class 11 Maths Chapter 7 Permutations and combinations
NCERT Solutions For Class 11 Maths Chapter 8 Binomial Theorem
NCERT Solutions For Class 11 Maths Chapter 9 Sequences and series
NCERT Solutions For Class 11 Maths Chapter 10 Straight lines
NCERT Solutions For Class 11 Maths Chapter 11 Conic sections
NCERT Solutions For Class 11 Maths Chapter 12 Introduction to threedimensional geometry
NCERT Solutions For Class 11 Maths Chapter 13 Limits and derivatives
NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning
NCERT Solutions For Class 11 Maths Chapter 15 Statistics
NCERT Solutions For Class 11 Maths Chapter 16 Probability
NCERT Solutions for Class 12 All Subjects  NCERT Solutions for Class 10 All Subjects 
NCERT Solutions for Class 11 All Subjects  NCERT Solutions for Class 9 All Subjects 
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