NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 14.

NCERT Solutions for Class 11 Maths Chapter 16 Probability contains solutions for all problems in Exercise 16.3. Chapter 16 Probability of Class 11 Maths is categorised under the CBSE Syllabus for the academic year 2023-24. Class 11 students can solve the NCERT textbook problems and gain confidence to attain good marks in the board exam. Students can simply download NCERT Solutions of Class 11 Maths now and practise offline.

NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3

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Access Solutions for Class 11 Maths Chapter 16 Exercise 16.3

1. Which of the following cannot be a valid assignment of probabilities for outcomes of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇}?

Assignment

Solution:-

(a) Condition (i): Each of the numbers p(ω_i ) is positive and less than zero. Condition (ii): Sum of probabilities

0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the given assignment is valid.

b) Condition (i): Each of the numbers p(ω_i ) is positive and less than zero. Condition (ii): Sum of probabilities

= (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7)

= 7/7

= 1

Therefore, the given assignment is valid.

c) Condition (i): Each of the numbers p(ω_i ) is positive and less than zero. Condition (ii): Sum of probabilities

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7

= 2.8 > 1

Therefore, the 2^nd condition is not satisfied.

Which states that p(w_i) ≤ 1

So, the given assignment is not valid.

d) The conditions of the axiomatic approach don’t hold true in the given assignment, that is

1) Each of the numbers p(w_i) is less than zero but also negative.

To be true, each of the numbers p(w_i) should be less than zero and positive.

So, the assignment is not valid.

e) Condition (i): Each of the numbers p(ω_i ) is positive and less than zero. Condition (ii): Sum of probabilities

= (1/14) + (2/14) + (3/14) + (4/14) + (5/14) + (6/14) + (7/14)

= (28/14) ≥ 1

The second condition doesn’t hold true, so the assignment is not valid.

2. A coin is tossed twice, what is the probability that at least one tail occurs?

Solution:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

Here, the coin is tossed twice, then the sample space is S = (TT, HH, TH, HT)

∴ Number of possible outcomes n (S) = 4

Let A be the event of getting at least one tail

∴ n (A) = 3

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

P(A) = n(A)/n(S)

= ¾

3. A die is thrown, find the probability of the 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:-

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

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

∴n(S) = 6

(i) A prime number will appear,

Let us assume ‘A’ be the event of getting a prime number,

A = {2, 3, 5}

Then, n(A) = 3

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(A) = n(A)/n(S)

= 3/6

= ½

(ii) A number greater than or equal to 3 will appear.

Let us assume ‘B’ be the event of getting a number greater than or equal to 3.

B = {3, 4, 5, 6}

Then, n(B) = 4

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(B) = n(B)/n(S)

= 4/6

= 2/3

(iii) A number less than or equal to one will appear.

Let us assume ‘C’ be the event of getting a number less than or equal to 1.

C = {1}

Then, n (C) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(C) = n(C)/n(S)

= 1/6

(iv) A number more than 6 will appear.

Let us assume ‘D’ be the event of getting a number more than 6; then,

D = {0)}

Then, n (D) = 0

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(D) = n(D)/n(S)

= 0/6

= 0

(v) A number less than 6 will appear.

Let us assume ‘E’ be the event of getting a number less than 6; then,

E= (1, 2, 3, 4, 5)

Then, n (E) = 5

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(E) = n(E)/n(S)

= 5/6

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:-

From the question, it is given that there are 52 cards in the deck.

(a) Number of points in the sample space = 52 (given)

∴n(S) = 52

(b) Let us assume ‘A’ be the event of drawing an ace of spades.

A= 1

Then, n (A) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(A) = n(A)/n(S)

= 1/52

(c) Let us assume ‘B’ be the event of drawing an ace. There are four aces.

Then, n (B)= 4

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(B) = n(B)/n(S)

= 4/52

= 1/13

(d) Let us assume ‘C’ be the event of drawing a black card. There are 26 black cards.

Then, n (C) = 26

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(C) = n(C)/n(S)

= 26/52

= ½

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:-

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

So, the sample space 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)}

Then, n(S) = 12

(i) Let us assume ‘P’ be the event having the sum of numbers as 3.

P = {(1, 2)},

Then, n (P) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(P) = n(P)/n(S)

= 1/12

(ii) Let us assume ‘Q’ be the event having the sum of the number as 12.

Then Q = {(6, 6)}, n (Q) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(Q) = n(Q)/n(S)

= 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?

Solution:-

From the question, it is given that there are four men and six women on the city council.

Here, the total members in the council = 4 + 6 = 10,

Hence, the sample space has 10 points.

∴ n (S) = 10

The number of women is 6 … [given]

Let us assume ‘A’ be the event of selecting a woman.

Then n (A) = 6

P(Event) = Number of outcomes favourable to the event/Total number of possible outcomes

∴P(A) = n(A)/n(S)

= 6/10 … [Divide both numerator and denominators by 2]

= 3/5

7. A fair coin is tossed four times, and a person wins Rs 1 for each head and loses 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:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

But, now coin is tossed four times, so the possible sample space contains,

S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH,

TTTH, TTHT, THTT, HTTT, TTTT)

As per the condition given the question, a person will win or lose money depending upon the face of the coin, so

(i) For 4 heads = 1 + 1 + 1 + 1 = ₹ 4

So, he wins ₹ 4.

(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50

= 3 – 1.50

= ₹ 1.50

So, he will be winning ₹ 1.50.

(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50

= 2 – 3

= – ₹ 1

So, he will be losing ₹ 1.

(iv)For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50

= 1 – 4.50

= – ₹ 3.50

So, he will be losing Rs. 3.50.

(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50

= – ₹ 6

So, he will be losing Rs. 6.

Now, the sample space of amounts is

S= {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1, – 1, – 1, – 1, – 3.50, – 3.50, – 3.50, – 3.50, – 6}

Then, n (S) = 16

P (winning ₹ 4) = 1/16

P (winning ₹ 1.50) = 4/16 … [Divide both numerator and denominator by 4]

= ¼

P (winning ₹ 1) = 6/16 … [Divide both numerator and denominator by 2]

= 3/8

P (winning ₹ 3.50) = 4/16 … [Divide both numerator and denominator by 4]

= ¼

P (winning ₹ 6) = 1/16

= 3/8

8. Three coins are tossed once. Find the probability of getting

(i) 3 heads (ii) 2 heads (iii) at least 2 heads

(iv) at most 2 heads (v) no head (vi) 3 tails

(vii) Exactly two tails (viii) no tail (ix) at most two tails

Solution:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

But, now three coins are tossed, so the possible sample space contains

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

Where s is sample space and here n(S) = 8

(i) 3 heads

Let us assume ‘A’ be the event of getting 3 heads.

n(A)= 1

∴P(A) = n(A)/n(S)

= 1/8

(ii) 2 heads

Let us assume ‘B’ be the event of getting 2 heads.

n (A) = 3

∴P(B) = n(B)/n(S)

= 3/8

(iii) at least 2 heads

Let us assume ‘C’ be the event of getting at least 2 heads.

n(C) = 4

∴P(C) = n(C)/n(S)

= 4/8

= ½

(iv) at most 2 heads

Let us assume ‘D’ be the event of getting at most 2 heads.

n(D) = 7

∴P(D) = n(D)/n(S)

= 7/8

(v) no head

Let us assume ‘E’ be the event of getting no heads.

n(E) = 1

∴P(E) = n(E)/n(S)

= 1/8

(vi) 3 tails

Let us assume ‘F’ be the event of getting 3 tails.

n(F) = 1

∴P(F) = n(F)/n(S)

= 1/8

(vii) Exactly two tails

Let us assume ‘G’ be the event of getting exactly 2 tails.

n(G) = 3

∴P(G) = n(G)/n(S)

= 3/8

(viii) no tail

Let us assume ‘H’ be the event of getting no tails.

n(H) = 1

∴P(H) = n(H)/n(S)

= 1/8

(ix) at most two tails

Let us assume ‘I’ be the event of getting at most 2 tails.

n(I) = 7

∴P(I) = n(I)/n(S)

= 7/8

9. If 2/11 is the probability of an event, what is the probability of the event ‘not A’?

Solution:-

From the question, it is given that 2/11 is the probability of an event A.

i.e., P (A) = 2/11

Then,

P (not A) = 1 – P (A)

= 1 – (2/11)

= (11 – 2)/11

= 9/11

10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that the letter is (i) a vowel (ii) a consonant

Solution:-

The word given in the question is ‘ASSASSINATION’.

Total letters in the given word = 13

Number of vowels in the given word = 6

Number of consonants in the given word = 7

Then, the sample space n(S) = 13

(i) a vowel

Let us assume ‘A’ be the event of selecting a vowel.

n(A) = 6

∴P(A) = n(A)/n(S)

= 6/13

(ii) Let us assume ‘B’ be the event of selecting the consonant.

n(B)= 7

∴P(B) = n(B)/n(S)

= 7/13

11. In a lottery, a person chooses 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: The order of the numbers is not important.]

Solution:-

From the question, it is given that

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

12. Check whether the following probabilities P(A) and P(B) are consistently defined

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

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

Solution:-

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

P(A ∩ B) > P(A)

Therefore, the given probabilities are not consistently defined.

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

Then,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.8 = 0.5 + 0.4 – P(A ∩ B)

Transposing – P(A ∩ B) to LHS, and it becomes P(A ∩ B) and 0.8 to RHS, and it becomes – 0.8.

P(A ∩ B) = 0.9 – 0.8

= 0.1

Therefore, P(A ∩ B) < P(A) and P(A ∩ B) < P(B)

So, the given probabilities are consistently defined.

13. Fill in the blanks in the following table.

Solution:-

From the given table,

(i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = ?

We know that,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (1/3) + (1/5) – (1/15)

= ((5 + 3)/15) – (1/15)

= (8/15) – (1/15)

= (8 – 1)/15

= 7/15

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

Then,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.6 = 0.35 + P(B) – 0.25

Transposing – 0.25, 0.35 to LHS, and it becomes 0.25 and – 0.35.

P(B) = 0.6 + 0.25 – 0.35

= 0.5

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

Then,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

0.7 = 0.5 + 0.35 – P(A ∩ B)

Transposing – P(A ∩ B) to LHS, and it becomes P(A ∩ B) and 0.7 to RHS, and it becomes – 0.7.

P(A ∩ B) = 0.85 – 0.7

= 0.15

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

Solution:-

From the question, it is given that

P(A) = 5/3 and P(B) = 1/5

Then, P(A or B), if A and B are mutually exclusive.

P(A∪B) or P(A or B) = P(A) + P(B)

= (3/5) + (1/5)

= 4/5

15. If E and F are events such that P(E) = ¼ , P(F) = ½ and P(E and F) = 1/8, find

(i) P(E or F), (ii) P(not E and not F)

Solution:-

From the question, we have P(E) = ¼ , P(F) = ½ and P(E ∩ F) = 1/8

(i) P(E or F), i.e., P(E∪F) = P(E) + P(F) – P(E ∩ F)

= ¼ + ½ – (1/8)

= 5/8

= 1 – (5/8)

= (8 – 5)/8

= 3/8

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:-

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:-

From the question, it is given that P(A) = 0.42, P(B) = 0.48 and 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) P(A not B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.42 + 0.48 – 0.16

= 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.

Solution:-

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?

Solution:-

Let us assume the probability of a randomly chosen student passing the first examination is 0.8 be P(A).

And also, assume the probability of passing the second examination is 0.7 be P(B).

Then,

P(A∪B) is the probability of passing at least one of the examinations.

Now,

P(A∪B) = 0.95 , P(A)=0.8, P(B)=0.7

∴ P(A∪B) = P(A) + P(B) – P(A∩B)

0.95 = 0.8 + 0.7 – P(A∩B)

Transposing – P(A ∩ B) to LHS, and it becomes P(A ∩ B) and 0.95 to RHS, and it becomes

– 0.95

P(A∩B) = 1.5 – 0.95

= 0.55

Hence, 0.55 is the probability that the student will pass both examinations.

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 us assume the probability of passing the English examination is 0.75 be P(A).

And also, assume the probability of passing the Hindi examination is P(B).

Here, given P(A) = 0.75, P(A∩B) – 0.5, P(A^I∩B^I) = 0.1

We know that, P(A^I∩B^I) = 1 – P(A∪B)

Then, P(A∪B) = 1 – P(A^I∩B^I)

= 1 – 0.1

= 0.9

∴ P(A∪B) = P(A) + P(B) – P(A∩B)

0.9 = 0.75 + P(B) – 0.5

Transposing 0.75, – 0.5 to LHS, and it becomes – 0.75, 0.5.

P(B) = 0.9 + 0.5 – 0.75

= 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.

(ii) The student has opted for neither NCC nor NSS.

(iii) The student has opted for NSS but not NCC.

Solution:-

From the question, it is given that

The total number of students in the class = 60

Thus, the sample space consists of n(S) = 60

Let us assume that the students opted for NCC to be ‘A’.

And also assume that the students opted for NSS to be ‘B’.

So, n(A) = 30, n(B) = 32 , n(A∩B) = 24

We know that, P(A) = n(A)/n(S)

= 30/60

= ½

P(B) = n(B)/n(S)

= 32/60

= 8/15

P(A∩B) = n(A∩B)/n(S)

= 24/60

= 2/5

Therefore, P(A∪B) = P(A) + P(B) – P(A∩B)

(i) The student opted for NCC or NSS.

P (A or B) = P(A) + P(B) –P(A and B)

P(A∪B) = P(A) + P(B) – P(A∩B)

= ½ + (8/15) – (2/5)

= 19/30

(ii) P(student opted neither for NCC nor NSS)

P(not A and not B) = P(A^I∩B^I)

We know that, P(A^I∩B^I) = 1 – P(A∪B)

= 1 – (19/30)

= 11/30

(iii) P(student opted for NSS but not NCC)

n(B – A) = n(B) – n (A∩B)

⇒ 32 – 24 = 8

The probability that the selected student has opted for NSS and not NCC is

= (8/60) = 2/15