NCERT Solutions for Class 11 Maths Chapter 16 – Probability

ANSWER:

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.

When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}

The number of elements in this sample space is 6 × 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)}

ANSWER:

When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).

When a coin is tossed four times, the total number of possible outcomes is 2⁴ = 16

Thus, when a coin is tossed four times, the sample space is given by:

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

ANSWER:

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.

Thus, when a coin is tossed and a die is thrown, the sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

ANSWER:

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.

Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T}

ANSWER:

Let us denote 2 boys and 2 girls in room X as B₁, B₂ and G₁, G₂ respectively. Let us denote 1 boy and 3 girls in room Y as B₃, and G₃, G₄, G₅ respectively.

Accordingly, the required sample space is given by S = {XB₁, XB₂, XG₁, XG₂, YB₃, YG₃, YG₄, YG₅}

ANSWER:

A die has six faces that are numbered from 1 to 6, with one number on each face.

Let us denote the red, white, and blue dices as R, W, and B respectively.

Accordingly, when a die is selected and then rolled, the sample space is given by

S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

ANSWER:

(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}

(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}

ANSWER:

It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.

When two balls are drawn at random in succession without replacement, the sample space is given by

S = {RW, WR, WW}

ANSWER:

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.

Thus, in the given experiment, the sample space is given by

S = {HH, HT, T1, T2, T3, T4, T5, T6}

ANSWER:

3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-defective (N).

The sample space of this experiment is given by

S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

ANSWER:

When a coin is tossed, the possible outcomes are head (H) and tail (T).

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.

Thus, the sample space of this experiment is given by:

S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

ANSWER:

If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.

Thus, the sample space of this experiment is given by 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)}

ANSWER:

A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.

A coin has two faces: head (H) and tail (T).

Hence, the sample space of this experiment is given by:

S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

ANSWER:

The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R₁, R₂ and the 3 black balls as B₁, B₂, and B₃.

The sample space of this experiment is given by

S = {TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}

ANSWER:

In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.

Hence, the sample space of this experiment is given by

S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …}

ANSWER:

When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.

Accordingly:

(i) A = {1, 2, 3, 4, 5, 6}

(ii) B = Φ

(iii) C = {3, 6}

(iv) D = {1, 2, 3}

(v) E = {6}

(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ

B ∪ C = {3, 6}, E ∩ F = {6}

D ∩ E =Φ, A – C = {1, 2, 4, 5}

D – E = {1, 2, 3}, 

ANSWER:

When a pair of dice is rolled, the sample space is given by

It is observed that

A ∩ B =Φ

B ∩ C =Φ

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

ANSWER:

When three coins are tossed, the sample space is given by

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Accordingly,

A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

We now observe that

A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ

B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ

C ∩ D = Φ

(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.

(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

ANSWER:

When three coins are tossed, the sample space is given by

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(i) Two events that are mutually exclusive can be

A: getting no heads and B: getting no tails

This is because sets A = {TTT} and B = {HHH} are disjoint.

(ii) Three events that are mutually exclusive and exhaustive can be

A: getting no heads

B: getting exactly one head

C: getting at least two heads

i.e.,

A = {TTT}

B = {HTT, THT, TTH}

C = {HHH, HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S

(iii) Two events that are not mutually exclusive can be

A: getting three heads

B: getting at least 2 heads

i.e.,

A = {HHH}

B = {HHH, HHT, HTH, THH}

This is because A ∩ B = {HHH} ≠ Φ

(iv) Two events which are mutually exclusive but not exhaustive can be

A: getting exactly one head

B: getting exactly one tail

That is

A = {HTT, THT, TTH}

B = {HHT, HTH, THH}

It is because, A ∩ B =Φ, but A ∪ B ≠ S

(v) Three events that are mutually exclusive but not exhaustive can be

A: getting exactly three heads

B: getting one head and two tails

C: getting one tail and two heads

i.e.,

A = {HHH}

B = {HTT, THT, TTH}

C = {HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ S

ANSWER:

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

(vii) B and C = B ∩ C 

ANSWER:

(i) It is observed that A ∩ B = Φ

∴ A and B are mutually exclusive.

Thus, the given statement is true.

(ii) It is observed that A ∩ B = Φ and A ∪ B = S

∴ A and B are mutually exclusive and exhaustive.

Thus, the given statement is true.

(iii) It is observed that

Thus, the given statement is true.

(iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠Φ

∴A and C are not mutually exclusive.

Thus, the given statement is false.

(v) 

∴A and are not mutually exclusive.

Thus, the given statement is false.

(vi) It is observed that;

However,

Therefore, events are not mutually exclusive and exhaustive.

Thus, the given statement is false.

ANSWER:

When a coin is tossed twice, the sample space is given by

S = {HH, HT, TH, TT}

Let A be the event of the occurrence of at least one tail.

Accordingly, A = {HT, TH, TT}

ANSWER:

The sample space of the given experiment is given by

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

(i) Let A be the event of the occurrence of a prime number.

Accordingly, A = {2, 3, 5}

(ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C = {1}

(iv) Let D be the event of the occurrence of a number greater than 6.

Accordingly, D = Φ

(v) Let E be the event of the occurrence of a number less than 6.

Accordingly, E = {1, 2, 3, 4, 5}

ANSWER:

(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements.

Therefore, there are 52 points in the sample space.

(b) Let A be the event in which the card drawn is an ace of spades.

Accordingly, n(A) = 1

(c) (i)Let E be the event in which the card drawn is an ace.

Since there are 4 aces in a pack of 52 cards, n(E) = 4

(ii)Let F be the event in which the card drawn is black.

Since there are 26 black cards in a pack of 52 cards, n(F) = 26

ANSWER:

Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by

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

Accordingly, n(S) = 12

(i) Let A be the event in which the sum of numbers that turn up is 3.

Accordingly, A = {(1, 2)}

(ii) Let B be the event in which the sum of numbers that turn up is 12.

Accordingly, B = {(6, 6)}

ANSWER:

There are four men and six women on the city council.

As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.

Let A be the event in which the selected council member is a woman.

Accordingly, n(A) = 6

ANSWER:

Since the coin is tossed four times, there can be a maximum of 4 heads or tails.

When 4 heads turns up, is the gain.

When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.

When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.

When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss.

When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.

There are 2⁴ = 16 elements in the sample space S, which is given by:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}

∴n(S) = 16

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

∴Probability (of winning Rs 4.00) =

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs.

∴Probability (of winning Rs 1.50) =

The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs.

∴Probability (of losing Re 1.00)

The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs.

Probability (of losing Rs 3.50) =

The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.

Probability (of losing Rs 6.00) =

ANSWER:

When three coins are tossed once, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴Accordingly, n(S) = 8

It is known that the probability of an event A is given by

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

∴P(B) =

(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}

∴P(C) =

(iii) Let D be the event of the occurrence of at least 2 heads.

Accordingly, D = {HHH, HHT, HTH, THH}

∴P(D) =

(iv) Let E be the event of the occurrence of at most 2 heads.

Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

∴P(E) =

(v) Let F be the event of the occurrence of no head.

Accordingly, F = {TTT}

∴P(F) =

(vi) Let G be the event of the occurrence of 3 tails.

Accordingly, G = {TTT}

∴P(G) =

(vii) Let H be the event of the occurrence of exactly 2 tails.

Accordingly, H = {HTT, THT, TTH}

∴P(H) =

(viii) Let I be the event of the occurrence of no tail.

Accordingly, I = {HHH}

∴P(I) =

(ix) Let J be the event of the occurrence of at most 2 tails.

Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}

∴P(J) =

ANSWER:

It is given that P(A) =.

Accordingly, P(not A) = 1 – P(A)

ANSWER:

There are 13 letters in the word ASSASSINATION.

∴Hence, n(S) = 13

(i) There are 6 vowels in the given word.

∴Probability (vowel) =

(ii) There are 7 consonants in the given word.

∴Probability (consonant) =

ANSWER:

Total number of ways in which one can choose six different numbers from 1 to 20 

Hence, there are 38760 combinations of 6 numbers.

Out of these combinations, one combination is already fixed by the lottery committee.

∴ Required probability of winning the prize in the game =

ANSWER:

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).

However, here, P(A ∩ B) > P(A).

Hence, P(A) and P(B) are not consistently defined.

(ii)P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).

Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B).

Hence, P(A) and P(B) are consistently defined.

ANSWER:

(i) Here,

We know that 

(ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6

We know that 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

⇒ P(B) = 0.5

(iii)Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7

We know that 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

⇒ P(A ∩ B) = 0.15

ANSWER:

Here, P(A) =, P(B) =

For mutually exclusive events A and B,

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

∴ P(A or B)

ANSWER:

Here, P(E) =, P(F) =, and P(E and F) =

(i) We know that P(E or F) = P(E) + P(F) – P(E and F)

∴P(E or F) =

(ii) From (i), P(E or F) = P (E ∪ F) =

ANSWER:

It is given that P (not E or not F) = 0.25

Thus, E and F are not mutually exclusive.

ANSWER:

It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16

(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58

(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52

(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)

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

ANSWER:

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

Accordingly, P(A) = 40% = = 

P(B) = 30% 

P(A and B) = 10% 

We know that P(A or B) = P(A) + P(B) – P(A and B)

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

ANSWER:

Let A and B be the events of passing first and second examinations respectively.

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

We know that P(A or B) = P(A) + P(B) – P(A and B)

∴0.95 = 0.8 + 0.7 – P(A and B)

⇒ P(A and B) = 0.8 + 0.7 – 0.95 = 0.55

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

ANSWER:

Let A and B be the events of passing English and Hindi examinations respectively.

Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e.,

P(A) = 0.75

We know that P(A or B) = P(A) + P(B) – P(A and B)

∴0.9 = 0.75 + P(B) – 0.5

⇒ P(B) = 0.9 – 0.75 + 0.5

⇒ P(B) = 0.65

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

ANSWER:

Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS.

Total number of students = 60

Number of students who have opted for NCC = 30

∴ P(A) =

Number of students who have opted for NSS = 32

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

(i) We know that P(A or B) = P(A) + P(B) – P(A and B)

Thus, the probability that the selected student has opted for NCC or NSS is.

(ii)

Thus, the probability that the selected students has neither opted for NCC nor NSS is.

(iii) The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8

Thus, the probability that the selected student has opted for NSS but not for NCC = 

ANSWER:

Number of ways of drawing 4 cards from 52 cards =

In a deck of 52 cards, there are 13 diamonds and 13 spades.

∴Number of ways of drawing 3 diamonds and one spade = 

Thus, the probability of obtaining 3 diamonds and one spade =

ANSWER:

Total number of faces = 6

(i) Number faces with number ‘2’ = 3

(ii) P (1 or 3) = P (not 2) = 1 − P (2) 

(iii) Number of faces with number ‘3’ = 1

ANSWER:

Total number of tickets sold = 10,000

Number prizes awarded = 10

(i) If we buy one ticket, then

P (getting a prize) = 

∴P (not getting a prize) =

(ii) If we buy two tickets, then

Number of tickets not awarded = 10,000 − 10 = 9990

P (not getting a prize) =

(iii) If we buy 10 tickets, then

P (not getting a prize) =

ANSWER:

My friend and I are among the 100 students.

Total number of ways of selecting 2 students out of 100 students =

(a) The two of us will enter the same section if both of us are among 40 students or among 60 students.

∴ Number of ways in which both of us enter the same section =

∴ Probability that both of us enter the same section

(b) P(we enter different sections)

= 1 − P(we enter the same section)

=

ANSWER:

Let L₁, L₂, L₃ be three letters and E₁, E₂, and E₃ be their corresponding envelops respectively.

There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

There are 4 ways in which at least one letter is inserted in a proper envelope.

Thus, the required probability is.

ANSWER:

It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35

(i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B)

∴P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88

(ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law]

∴P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12

(iii) P(A ∩ B′) = P(A) − P(A ∩ B)

= 0.54 − 0.35

= 0.19

(iv) We know that

ANSWER:

Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age.

Accordingly, P(E) = and P(F) =

Since there is only one male who is over 35 years of age,

We know that 

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

ANSWER:

(i) When the digits are repeated

Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.

The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed.

∴Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 − 1

= 250 − 1 = 249

[In this case, 5000 can not be counted; so 1 is subtracted]

A number is divisible by 5 if the digit at its units place is either 0 or 5.

∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 − 1 = 100 − 1 = 99

Thus, the probability of forming a number divisible by 5 when the digits are repeated is.

(ii) When repetition of digits is not allowed

The thousands place can be filled with either of the two digits 5 or 7.

The remaining 3 places can be filled with any of the remaining 4 digits.

∴Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2

= 48

When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴Here, number of 4-digit numbers starting with 5 and divisible by 5

= 3 × 2 = 6

When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴ Here, number of 4-digit numbers starting with 7 and divisible by 5

= 1 × 2 × 3 × 2 = 12

∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is.

ANSWER:

The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.

Number of ways of selecting 4 different digits out of the 10 digits =

Now, each combination of 4 different digits can be arranged in ways.

∴Number of four digits with no repetitions =

There is only one number that can open the suitcase.

Thus, the required probability is.