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**NCERT Exemplar Class 11 Maths solutions chapter 16** discusses Probability and its applications in real life. Students who find the problems of Probability difficult or want to practice for the exams more can refer to NCERT Exemplar solutions for Class 11 Maths chapter 16 and follow the given steps as mentioned by our experts. The solutions are given in a step-by-step manner which enables the students to understand the solution easily. The students can practice and understand the steps demonstrated in the NCERT Exemplar Class 11 Maths solutions chapter 16 for their CBSE board examinations.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

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This Story also Contains

- NCERT Exemplar Class 11 Maths Solutions Chapter 16: Exercise: 1.3
- More About NCERT Exemplar Class 11 Maths Chapter 16
- List of Topics and Sub-topics in NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability:
- NCERT Solutions for Class 11 Mathematics Chapters
- Important Topics to Cover in NCERT Exemplar Class 11 Maths Solutions Chapter 16

Question:1

Answer:

ALGORITHM ……. (given word)

Total no. of letters = 9

Thus, total no. of words = 9!

Thus, n(s) = 9!

Considering ‘GOR’ as one group –

A L GOR I T H M

↓ ↓ ↓ ↓ ↓ ↓ ↓

1 2 3 4 5 6 7

Thus, no. of letters = 7

Now, if the GOR group remains together, then the order = 7!

Thus, n (E) = 7!

Now, we know that,

Required probability = No. of favorable outcomes/ Total no. of outcomes

= n (E) / n(S) …..[Since n! = n x (n – 1) x (n – 2)…1]

= 7! / 9!

= 7! / 9 x 8 x 7!

= 1 / 72

Question:2

Answer:

Given: total no. of employees = 6

They can be arranged in 6 ways,

Thus, n(S) = 6!

= 6 x 5 x 4 x 3 x 1

= 720

Now, there are 5 different ways to select two adjacent desks for married couples –

(1,2), (2,3), (3,4), (4,5), (5,6)

They can be arranged in 2! ways in the two desks & the other persons can be arranged in 4! Ways

Thus, the no. of ways = 5 x 2! X 4!

= 5 x 2 x 1 x 4 x 3 x 2 x 1

= 240

Thus,

The no. of ways in which married couples occupy non- adjacent desks

= 6! – 240

= 720 – 240

= 480

= n (E)

Required probability = No. of favorable outcomes/ Total no. of outcomes

= n (E) / n(S)

= 480 / 720

= 2 / 3

Question:3

Answer:

Given: we have integers 1, 2, …….. , 1000

No. of outcomes, n(S) = 1000

No. of the integers that are multiples of 2 –

2, 4, 6, 8, …… , 1000.

Let us consider ‘p’ as the no. of integers,

Now, ap = a + (p – 1)d

On substituting the values, we get,

2 + (p – 1)2 = 1000

2 + 2p – 2 = 1000

Thus, p = 1000/ 2

Thus, p = 500

Thus, no. of the integers that are multiples of 2 = 500

No. of the integers that are multiples of 9 –

9, 18, 27, 35, …… , 999.

Let us consider ‘n’ as the no. of integers,

Now, an = a + (n – 1)d

On substituting the values, we get,

9 + (n – 1)9 = 999

9 + 9n – 9 = 999

Thus, n = 999/ 9

Thus, n = 111

Thus, no. of the integers that are multiples of 9 = 111

Now, let m be the no. of multiples common for both 2 & 9, viz. 18, 36, ……., 990.

Thus, the mth term will be 990

Now, am = a + (m – 1)d

We know that, a = 2 & d = 9

Substituting the respective values, we get,

18 + (m – 1)18 = 990

18 + 18m – 18 = 990

Thus, m = 990/18

Thus, m = 55

Now, the no. of multiples of 2 or 9 will be,

No. of multiples of 2 + no. of multiples of 9 – No. of multiples of both 2 & 9

= 500 + 111 – 55

= 556

= n(E)

Required probability = No. of favorable outcomes/ Total no. of outcomes

= n(E) / n(S)

= 556 / 1000

= 0.556

Question:4

Answer:

We know that, the no. of outcomes when a die is thrown is 6

2 appear on the kth roll of the die ……… (given)

Thus, the first (k – 1)th roll has 5 outcomes each

Thus, no. of outcomes = 5^{k-1}

Let us consider that 2 come before the kth roll of the die and not after that.

Thus,

In the first roll, no. of ways in which 2 appears will be = 1 outcome

In the second roll, no. of ways in which 2 appears will be = 5 x 1 outcome

……. (since the first roll doesn’t result in 2)

In the third roll, no. of ways in which 2 appears will be = 5 x 5 x 1 outcome

……. (since the first two rolls doesn’t result in 2)

In the (k – 1)^{th} roll, no. of ways in which 2 appear will be = [5 x 5 x 1 ….. (k – 1)] outcome

= 5^{k-1}

Now, the possibility of 2 appearing before kth roll = 1 + 5 + 5^{2} + 5^{3} + …… + 5^{k-1}

Now,

Thus, here, a = 1 & r = 5/1 = 5 >1

Thus,

= 1 x (5^{k} – 1) / 5 – 1

= 5^{k} – 1 / 4

Question:5

Answer:

Probability of odd nos. = 2 x (probability of even no.) ……….. (given)

Thus, P (Odd) = 2 x P (Even)

P(Odd) + P(Even) = 1

2P (Even) + P (Even) = 1

3P (Even) = 1

Thus, P (Even) = 1 / 3

Thus, P (Odd) = 1 – 1/3

= 3 – 1/ 3

= 2 / 3

Total no. occurring on a single roll = 6

& 4, 5 & 6 are the nos. greater than 3

Let P(no. greater than 3) = P (G)

= P (no. is 4, 5 or 6)

Here, 4 & 6 – Even & 5 – Odd

Thus, P (G) = 2 x P (Even) x P (Odd)

= 2 x 1/3 x 2/3

= 4/9

Therefore, 4/9 is the required probability.

Question:6

Answer:

Let us consider that,

E1 be the element that a family owns a colour TV & E_{2 }be the event that a family owns black & white TV.

Now P(E_{1}) = 0.87 & P(E_{2}) = 0.36 …….. (given)

& P (E_{1} ∩ E_{2}) = 0.30

To find: probability that the family owns either anyone or both kinds of sets.

Now, by the general rule –

P (A U B) = P(A) + P(B) – P(A ∩ B)

We have,

P (E_{1} U E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= 0.87 + 0.36 – 0.30

= 1.23 – 0.30

= 0.93

Therefore, 0.93 is the required probability

Question:7

Answer:

P(A) = 0.35 & P(B) = 0.45 ….. (given)

P(A ∩ B) = 0 ……. (since A & B are mutually exclusive)

P(A’)

Now, we know that,

P (A) + P (A’) = 1

0.35 + P (A’) = 1

P(A’) = 1 – 0.35

P (A’) = 0.65

P (B’)

Now, we know that,

P (B) + P (B’) = 1

0.45 + P (B’) = 1

P (B’) = 1 – 0.45

P (B’) = 0.55

P (a u b)

Now, we know that,

P (A U B) = P(A) + P(B) – P(A ∩ B)

P (A U B) = 0.35 + 0.45 – 0

P (A U B) = 0.80

P (A ∩ B)

Since A & B are mutually exclusive events,

Thus, P (A ∩ B) = 0

P (A ∩ B’)

P (A ∩ B’) = P (A) - P (A ∩ B)

= 0.35 – 0

= 0.35

P (A’ ∩ B’)

P (A’ ∩ B’) = P (A U B)’

= 1 – P (A U B)

= 1 – 0.8 …… [from (c)]

= 0.2

Question:8

Answer:

Given:

P (E_{1}) = 0.15

P (E_{2}) = 0.20

P (E_{3}) = 0.31

P (E_{4}) = 0.26

P (E_{5}) = 0.08

Let us consider that –

E_{1 }→ Event that surgeries are rated as very complex

E_{2} → Event that surgeries are rated as complex

E_{3} → Event that surgeries are rated as routine

E_{4} → Event that surgeries are rated as simple

E_{5} → Event that surgeries are rated as very simple

P (complex or very complex) = P (E

_{1}or E_{2})

= P (E_{1} U E_{2})

Now, by the general rule –

P (A U B) = P(A) + P(B) – P(A ∩ B)

We have,

P (E_{1 }U E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= 0.15 + 0.2 – 0

= 0.35

P (neither very complex nor very simple) = P (E

_{1}’ ∩ E_{5}’)

= 1 - P (E_{1} ∩ E_{5}) ….. (by complement rule)

= 1 – [P (E_{1}) + P (E_{5}) – P (E_{1} ∩ E_{5}) … (general addn rule)

= 1 – [0.15 + 0.08 – 0]

= 1 – 0.23

= 0.77

P (routine or complex) = P (E

_{3}∩ E_{2})

= P (E_{3}) + P (E_{2}) - P (E_{3} ∩ E_{2}) …… (by general addition rule)

= 0.31 + 0.2 – 0

= 0.51

P (routine or simple) = P (E

_{3}∩ E_{4})

= P (E_{3}) + P (E_{4}) - P (E_{3} ∩ E_{4}) …… (by general addition rule)

= 0.31 + 0.26 - 0

= 0.57

Question:9

Answer:

Given:

A is twice likely to be selected as B, P(A) = 2 P(B)

& C is twice likely to be selected as D, P(C) = 2 P(D)

It is given that B & C have about the same chance

Thus, P(B) = P(C)

Now, sum of all probabilities is 1,

Thus,

P(A) + P(B) + P(C) + P(D) = 1

P(A) + P(B) + P(B) + P(D) = 1

Thus,

P(A) + P(A)/2 + P(A)/2 + P(C)/2 = 1

[2 P(A) + P(A) + P(A) + P(B)] /2 = 1

4 P(A) + P(A) / 2 = 2

[8 P(A) + P(A)] / 2 = 2

9 P(A) = 4

P(A) = 4/9

Now, (a) P (C will be selected) = P (C)

= P (B)

= 4/9 x ½

= 2/9

(b) P (A will not be selected) = P (A’)

= 1 – P (A) ……. (by complement rule)

= 1 – 4/9

= 9-4/ 9

= 5 / 9

Question:10

Answer:

Given:

Sample Space (S) = John promoted, Rita promoted, Aslam promoted, Gurpreet promoted

Chances of John’s promotion is same as Gurpreet’s, P (E_{1}) = P (E_{4})

Rita’s chances of promotion as twice as john’s, P (E_{2}) = 2 P(E_{1})

& chances of Aslam’s promotion are four times that of John’s, P (E_{3}) = 4 P(E_{1})

Now, let us consider that,

E_{1 }→ events that John promoted

E_{2 }→ events that Rita promoted

E_{3} → events that Aslam promoted

E_{4} → events that Gurpreet promoted

Now, we know that,

Sum of all probabilities = 1

Thus, P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) = 1

P(E_{1}) + 2 P(E_{1}) + 4 P(E_{1}) + P(E_{1}) = 1

8 P(E_{1}) = 1

P (E_{1}) = 1/8

Now,

1. P (John promoted) = P (E_{1}) = 1/8

P (Rita promoted) = P (E_{2}) = 2P (E_{1})

= 2 x 1/8

= 1/4

P (Aslam promoted) = P (E_{3}) = 4P (E_{1})

= 4 x 1/8

= 1/2

P (Gurpreet promoted) = P (E_{4}) = P (E_{1})

= 1/8

2. A = John or Gurpreet promoted …….. (Given)

Thus, A = E_{1} U E_{2}

P (A) = P (E_{1} U E_{2})

= P (E_{1}) + P (E_{4}) – P (E_{1} ∩ E_{2}) ……. (general addition rule)

= P (E_{1}) + P (E_{1}) – 0

= 1/8 + 1/8

= 2/8

= 1/4

Question:11

Answer:

P (A ∩ B) = 0.07 ……. (given)

1. P (A) = 0.13 + 0.7 …… (by given Venn diagram)

= 0.20

2. P (B ∩ ) = P (B) – P (B ∩ C)

= 0.07 + 0.10 + 0.15 – 0.15

= 0.07 + 0.10

= 0.17

3. P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

P (A U B) = 0.20 + (0.07 +0.10 +0.15) – 0.07

P (A U B) = 0.20 + 0.25

P (A U B) = 0.45

4. P (A ∩) = P (A) – P (A ∩ B)

= 0.20 – 0.07

= 0.13

5. P (B ∩ C) = 0.15 …… (from venn diagram)

6. P (exactly one of the 3 occurs) = 0.13 + 0.10 + 0.28 = 0.51

Question:12

Answer:

Given:

One urn contains 2 black balls & 1 white ball

The other urn contains 1 black ball & 2 white balls

If one of the 2 urns is chosen then a ball is randomly chosen from urn & without replacing the first ball, the second ball is also chosen from the same urn.

Now, we know that,

Sample Space, S = {B_{1}B_{2}, B_{1}W, B_{2}W, B_{2}B_{1}, B_{2}W, WB_{1}, WB_{2}, W_{1}W_{2}, W_{1}B, W_{2}B, W_{2}B, W_{2}W_{1}, BW_{1}, BW_{2}}

Thus, total no. of sample space = 12

Now, if two black balls are chosen,

Favorable outcomes = B_{1}B_{2} & B_{2}B_{1}

& total no. of favorable outcomes = 2

Now,

Probability = no. of favorable outcomes / total no. of outcomes

= 2 /12 = 1 /6

Now, if two balls of opposite colours are chosen,

Favorable outcomes = B_{1}W, B_{2}W, WB_{1}, WB_{2}, W_{1}B, W_{2}B, BW_{1}, BW_{2}

& total no. of favorable outcomes = 8

Now,

Probability = no. of favorable outcomes / total no. of outcomes

= 8 /12 = 2 /3

Question:13

Answer:

Given:

No. of reds balls = 8

No. of white balls = 5

Thus, total no. of balls, n = 13

Now, 3 balls are drawn at random, thus,

r = 3

Thus,

n (S) = ^{n}C_{r}

= ^{13}C_{3}

If all balls are white,

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

= no. of favorable outcomes/ sample space

Now, total white balls are = 5

Thus,

P (all the three balls are white) = ^{5}C_{3} / ^{13}C_{3}

= 5 /143

All three balls are red,

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

= no. of favorable outcomes/ sample space

Now, total red balls are = 8

Thus,

P (all the three balls are white) = ^{8}C_{3} / ^{13}C_{3}

= 28 /143

One ball is red and two balls are white

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

= no. of favorable outcomes/ sample space

Now, total white balls are = 5

Thus,

P (One ball is red and two balls are white) = ^{8}C_{1} x ^{5}C_{2} / ^{13}C_{3}

= 40 /143

Question:14

Answer:

Given word: ASSASINATION

Total no. of letters in the word = 13

Viz., 3 A’s, 4 S’s, 2 I’s, 1 T & 1 O

No. of ways in which these letters can be arranged –

n(S) = 13! / 3! 4! 2! 2!

4 S’s come consecutively in ASSASINATION

Then the word becomes-

S | S | S | S | A | A | I | N | A | T | I | O | N |

No. of letters = 1 + 9 = 10Now,

Thus,

n(E) = 10! / 3! 2! 2!

Now,

Required probability =

= 10! / 3! 2! 2! x 3! 4! 2! 2! / 13!

= 10! x 4!/ 13 x 12 x 11 x 10!

= 2 /143

2 I’s & 2 N’s come together

Then the word becomes-

I | I | N | N | A | S | S | A | S | S | A | T | O |

No. of letters = 1 + 9 = 10Now,

Thus,

n(E) = 4! / 2! 2! x 10!/ 3! 4!

Now,

Required probability =

= 10! 4! / 3! 4! 2! 2! x 3! 4! 2! 2! / 13!

= 10! x 4!/ 13 x 12 x 11 x 10!

= 2 /143

All A’s are not coming together

Then the word becomes-

A | A | A | S | S | S | S | I | N | T | I | O | N |

Now,

No. of letters = 1 + 10 = 11

Thus,

When all A’s come together no. of words = 11! / 4! 2! 2!

Now,

probability =

= 11! / 4! 2! 2! x 3! 4! 2! 2! / 13!

= 11! x 3!/ 13 x 12 x 11!

= 1 / 26

Now,

P (All A’s don’t come together) = 1 – P (all A’s comes together)

= 1 – 1/26

= 25 / 26

No two A’s are coming together,

Then the word becomes-

S | S | S | S | I | N | T | I | O | N |

No. of ways of arranging except A = 10! / 4! 2! 2!Now,

There are 11 vacant places

Total no. of A’s in ASSASINATION = 3

Thus, the 3 A’s can be placed in ^{11}C_{3 }ways

= 11! / 3! (11 – 3)!

= 11! / 3! 8!

No. of ways when 2 A’s are not together

= 11! / 3! 8! x 10! / 4! 2! 2! x 3! 4! 2! 2!/ 13!

= 11! X 10 x 9 x 8! / 8! x 13 x 12 x 11!

= 10 x 9 / 13 x 12

= 15 / 26

Question:15

Answer:

We know that, in a deck,

Total no. of cards = 52

No. of kings = 4

No. of heart cards = 13

& total no. of red cards = 13 + 13 = 26

Thus, favorable outcomes = 4 + 13 + 26 – 13 – 2

= 28

Now,

Probability = no. of favorable outcomes / total no. of outcomes

= 28 / 52

= 7 / 13

Question:16

Answer:

Given data:

S = {e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}, e_{7}, e_{8}, e_{9}}

A = {e_{1}, e_{5}, e_{8}}

B = { e_{2}, e_{5 }e_{8}, e_{9}}

P (e_{1}) = P (e_{2}) = 0.08

P (e_{3}) = P (e_{4}) = P (e_{5}) = 0.1

P (e_{6}) = P (e_{7}) = 0.2

P (e_{8}) = P (e_{9}) = 0.07

(a) P (A), P (B) & P (A ∩ B)

A = {e_{1}, e_{5}, e_{8}} …….. (given)

Thus,

P (A) = P (e_{1}) + P (e_{5}) + P (e_{8})

= 0.08 + 0.1 + 0.07

= 0.25

Now, B = { e_{2}, e_{5} e_{8}, e_{9}} ……. (given)

P (B) = P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9})

Now, P (A ∩ B)

A ∩ B = {e_{5}, e_{8}}

Thus, P (A ∩ B) = P (e_{5}) + P (e_{8})

= 0.1 + 0.07

= 0.17

(b) P (A U B)

P (A) = 0.25

P (B) = 0.32

P (A ∩ B) = 0.17

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

= 0.25 + 0.32 – 0.17

= 0.40

(c) A = {e_{1}, e_{5}, e_{8}}

B = { e_{2}, e_{5} e_{8}, e_{9}} ……. (given)

Thus, A U B = {e_{1}, e_{2}, e_{5}, e_{8}, e_{9}}

Thus, P (A U B) = P (e_{1}) + P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9})

= 0.08 + 0.08 + 0.1 + 0.07 + 0.07

= 0.40

(d) P ()

P () = 1 – 0.32

= 0.68

Now, we have,

B = { e_{2}, e_{5 }e_{8}, e_{9}}

Thus, = { e_{1}, e_{3}, e_{4}, e_{6}, e_{7}}

P () = P (e_{1}) + P (e_{3}) + P (e_{4}) + P (e_{6}) + P (e_{7})

= 0.08 + 0.1 + 0.1 + 0.2 +0.2

= 0.68

Question:17

(a) We know that, the possible outcomes of a fair die are-

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

Thus, total no. of outcomes = 6

Here, 1, 3 & 5 are odd nos.

Thus, favorable outcomes = 3

Probability = no. of favorable outcomes/ total no. of outcomes

= 3 / 6

= 1/2

(b) When a fair coin is tossed twice,

S = {HH, HT, TH, TT}

Thus, total n, of outcomes = 4

For at least one head to appear, the possible cases will be – HH, HT, TH

Thus, favorable outcomes = 3

Now,

Probability = no. of favorable outcomes/ total no. of outcomes

= 3/ 4

When we roll a pair of dice,

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)}

Thus, n(S) = 36

For sum to be 6, the possible outcomes will be –

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

Thus, favorable outcomes = 5

Probability = no. of favorable outcomes/ total no. of outcomes

= 5/ 36

Question:18

In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is

A. 1/7

B. 2/7

C. 3/7

D. none of these

Answer:

B

(i) A non-leap year cofltains 365 days. So, on dividing it by 7, we get 52 weeks and 1 more day.

So, since 52 weeks are there, it means 52 Tuesdays will also be there necessarily with probability I and 1 more day may be either Sun. or Mon. or Tue. or Wed. or Thur. or Fri. or Sat.

So, to get 53 Tuesdays, we have to select one more Tuesday from these 7 possibilities with Probability 1/7.

Therefore, probability of having 53 Tuesdays or 53 wednesdat in a non-leap year

Question:19

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

A.

B.

C.

D.

Answer:

The set has 3 consecutive nos. from 1 to 20,

Thus, it is – (1,2,3), (2,3,4), (3,4,5), …….. , (18,19,20)

Now, if we consider 3 nos. as a single digit, there will be 18 nos.

Now, choosing 3 nos. out of 20 can be done in ^{20}C_{3} ways

This, n(S) = ^{20}C_{3}

Required event – the 3 nos. chosen must be consecutive, thus,

P (nos. are consecutive) = 18/ ^{30}C_{3}

Now,

P (nos. that are not consecutive) = 1 – 3/190

= 190 – 3/ 190

= 187 / 190

Thus, option B is the correct answer.

Question:20

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours

A. 29/52

B. 1/2

C. 26/51

D. 27/51

Answer:

There are 26 red cards & 26 black cards in a pack of 52 cards.

Now,

2 cards are accidently dropped …….. (given)

Thus,

Probability of dropping a red card first = 26 /52

& probability of dropping a red card second = 26 / 51

…….. (since one card is already dropped, we are left with 51 cards)

Now,

Probability of dropping a black card first = 26 /52

Probability of dropping a black card second= 26 /51

Thus,

P (both cards are of diff colour) = 26/52 x 26/51 + 26/52 + 26/51

= 2 x 26/52 x 26/51

= 26 /51

Thus, option C is the correct answer.

Question:21

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is

A. 1/3

B. 1/6

C. 2/7

D. 1/2

Answer:

7 persons are to be seated in a row ….. (given)

Let us consider 2 persons as 1 group since 2 person sit next to each other

Thus, we have to arrange 6 persons,

Thus, the no. of arrangement = 2! X 6!

Thus, total no. of arrangement of 7 persons = 7!

Now,

Probability = no. of favorable outcomes/ total no. of outcomes

= 2! X 6! / 7!

= 2 x 1 x 6! / 7 x 6!

= 2/ 7

Thus, option C is the correct answer.

Question:22

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is

A. 1/5

B. 4/5

C. 1/30

D. 5/9

Answer:

Given digits: 0, 2, 3, 5

Now, we now that if a no. is divisible by 5, then the digit in the units place should be either 0 or 5

Now, if units place is 0, then

3 | 2 | 1 | 1 |

Now, the first three places can be filled in 3! Ways

= 3 x 2 x 1 x 1

= 6

Now, if units place is 5, then,

2 | 1 | 1 | 1 |

Here, we can fill the first two places in 2 ways & the 2^{nd }and 3^{rd} place in 2! Ways

= 2 x 2 x 1 x 1

= 4

Thus, the total no. of ways = 6 + 4

= 10

= n (E)

Now, total no. of ways to arrange the given digits to form 4 – digits without repetition is 3 x 3 x 2 x 1

= 18

Probability = no. of favorable outcomes/ total no. of outcomes

= 10 /18

= 5 /9

Thus, option D is the correct answer.

Question:23

If A and B are mutually exclusive events, then

A. P (A) ≤ P ()

B. P (A) ≥ P ()

C. P (A) < P ()

D. none of these

Answer:

A & B are mutually exclusive events …… (given)

Thus, P (A ∩ B) = 0

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

Thus, P (A U B) = P (A) + P (B) - P (A ∩ B)

P (A U B) = P (A) + P (B) – 0

P (A U B) = P (A) + P (B)

Now, for all events A, B

0 ≤P(A)≤1

Thus, P (A) + P (B) ≤ 1

P (A) ≤ 1 – P (B)

Thus, by complement rule,

P (A) ≤ P ()

Thus, option A is the correct answer.

Question:24

If P (A ∪ B) = P (A ∩ B) for any two events A and B, then

A. P (A) = P

B. (B) P (A) > P (B)

C. P (A) < P (B)

D. none of these

Answer:

P (A ∩B) = P (A U B)

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

P (A ∩ B) = P (A) + P (B) – P (A ∩ B)

Thus, [P (A) – P (A ∩ B)] + [P (B) – P (A ∩ B)] = 0

Now,

P (A) – P (A ∩ B) ≥ 0

& P (B) – P (A ∩ B) ≥ 0

P (A) – P (A ∩ B) = 0

P (B) – P (A ∩ B) = 0

Thus, P (A) = P (A ∩ B)

& P (B) = P (A ∩ B)

Therefore, we get

P(A) = P (B)

Thus, option A is the correct answer.

Question:25

6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

A. 1/432

B. 12/431

C. 1/132

D. none of these

Answer:

Let us consider group 1 where all girls sit together,

Thus,

G | G | G | G | G | G | 6 BOYS |

Total persons = 1 + 6

= 7

Now, total no. of arrangements in row of 7 = 7!

Also, the ways in which girls interchange their seats is 6!

Probability = no. of favorable outcomes/ total no. of outcomes

= 6! 7! / 12!

= 6 x 5 x 4 x 3 x 2 x 1 x 7! / 12 x 11 x 10 x 9 x 8 x 7!

= 6 x 5 x 4 x 3 x 2 / 12 x 11 x 10 x 9 x 8

= 1 / 2 x 11 x 2 x 3 x 2

= 1 / 132

Thus, option C is the correct answer.

Question:26

A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

A. 1/3

B. 4/11

C. 2/11

D. 3/11

Answer:

Total no. of alphabet in PROBABILITY = 11

& no. of vowels in it = 4, viz., (O, A, I, I)

Probability = no. of favorable outcomes/ total no. of outcomes

Now, P (letter is vowel) = 4 / 11

Thus, option B is the correct answer.

Question:27

If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

A. >0 . 5

B. 0.5

C. ≤ 0.5

D. 0

Answer:

Let us consider that-

E_{1} → event that A fails in the examination

& E_{2 }→ event that B fails in the examination

Thus,

P (E_{1}) = 0.2

P (E_{2}) = 0.3

To find: P (either E_{1} or E_{2} fails)

Thus, P (either E_{1} or E_{2 }fails) = P (E_{1}) + P (E_{2}) – P (E_{1} ∩ E_{2})

≤ P(E_{1}) + P (E_{2})

≤ 0.2 + 0.3

≤ 0.5

Thus, option C is the correct answer.

Question:28

The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P () + P () is

A. 0.4

B. 0.8

C. 1.2

D. 1.6

Answer:

Given:

P ( at least one of A or B occurs) = 0.6, thus, P (A U B) = 0.6

& P (A & B occurs simultaneously) = 0.2, thus, P (A ∩ B) = 0.2

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

Thus, P (A) + P (B) – 0.2 = 0.6

P (A) + P (B) = 0.6 + 0.2

P (A) + P (B) = 0.8

Now,

P (A) = 1 – P (A’) ……… by complement rule

Similarly, P (B) = 1 – P (B’)

Therefore,

1 – P (A’) + 1 – P (B’) = 0.8

2 – [P (A’) + P (B’)] = 0.8

2 – 0.8 = P (A’) + P (B’)

P (A’) + P (B’) = 1.2

Thus, option C is the correct answer.

Question:29

If M and N are any two events, the probability that at least one of them occurs is

A. P (M) + P (N) – 2 P (M ∩ N)

B. P (M) + P (N) – P (M ∩ N)

C. P (M) + P (N) + P (M ∩ N)

D. P (M) + P (N) + 2P (M ∩ N)

Answer:

M and N are two events ……. (given)

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

Thus, P (M U N) = P (M) + P (N) – P (M ∩ N)

Thus, option B is the correct answer.

Question:30

Answer:

Let us consider that,

E_{1 }→ event that person see the giraffe

E_{2} → event that person see the bear

We have,

P (E_{1}) = 0.72

& P (E_{2}) = 0.84

P (person will see giraffe as well as bear) = 0.52

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

P (E_{1} U E_{2}) = P (E_{1}) + P (E_{2}) – P (E_{1} ∩ E_{2})

= 0.72 + 0.84 – 0.52

= 1.04 viz. not possible

Thus, the given statement is false.

Question:31

Answer:

Let us consider that,

A → event that student pass examination

B → event that student get compartment

We have,

P (A) = 0.73

P (B) = 0.13

& P (A U B) = 0.96

To find: P (A ∩ B)

Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

P (A U B) = 0.73 + 0.13 – 0

= 0.86

but given P (A U B) = 0.96

Thus, the given statement is False.

Question:32

Answer:

Let us consider that –

A → event that typist will make 0 mistake

B → event that typist will make 1 mistake

C → event that typist will make 2 mistake

D → event that typist will make 3 mistake

E → event that typist will make 4 mistake

F → event that typist will make 5 mistake

We have,

P (A) = 0.12

P (B) = 0.25

P (C) = 0.36

P (D) = 0.14

P (E) = 0.08

P (F) = 0.11

Now,

Sum of all probabilities = 1

Thus,

P (A) + P (B) + P (C) + P (D) + P (E) + P (F)

= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11

= 1.06 viz. > 1

Thus, the given statement is False.

Question:33

Answer:

Let us consider that,

E_{1} → event that A is selected in engineering college

E_{2} → event that B is selected in engineering college

We have,

P (E_{1}) = 0.5

& P (E_{2}) = 0.7

P (E_{1} ∩ E_{2}) ≤ 0.3

Thus,

0.5 x P (E_{2}) ≤ 0.3

P (E_{2}) ≤ 0.6

But, we know that,

P (E_{2}) = 0.7

Thus, the given statement is False.

Question:34

Answer:

A & B are two events

A ∩ B A

P (A ∩ B) ≤ P (A)

Thus, the given statement is true.

Question:35

Answer:

Given data:

P (occurrence of event A) = 0.7

P (occurrence of event B) = 0.3

P (occurrence of both) = 0.4

Now,

P (A ∩ B) = P (A) x P (B)

P (A ∩ B) = 0.7 x 0.3

= 0.21

But, given

P (A ∩ B) = 0.4

Thus, the given statement is False.

Question:36

Answer:

We know that,

Probability of each student getting distinction in final examination is ≤ 1

Now, the given events are not related to the sample space

Thus, 1.2 may be the sum of their probabilities.

Thus, the given statement is true

Question:37

Answer:

Let us consider that,

A → event that home team will win football game

B → event that home team will tie football game

C → event that home team will lose football game

It is given that,

P (A) = 0.77

& P (B) = 0.08

To find: P (C)

Now, we know that,

Sum of all probabilities = 1

Thus,

P (A) + P (B) + P (C) = 1

0.77 + 0.08 + P (C) = 1

P (C) = 1 – (0.77 + 0.08)

= 0.15

Answer = 0.15

Question:38

Answer:

Given data:

P (e_{1}) = 0.1

P (e_{2}) = 0.5

P (e_{3}) = 0.1

Now, we know that,

Sum of all probabilities = 1

Thus,

P (e_{1}) + P (e_{2}) + P (e_{3}) + P (e_{4}) = 1

0.1+ 0.5 + 0.1 + P (e_{4}) = 1

P (e_{4}) = 1 – 0.7

Thus, P (e_{4}) = 0.3

Answer: 0.3

Question:39

Fill in the blanks

Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then is _________.

Answer:

Given data:

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

& E = {1, 3, 5}

To find:

Now,

= S – E {w:w ? S & w ?E}

= {2, 4, 6}

Answer: {2, 4, 6}

Question:40

Answer:

Given data:

P (A) = 0.3

P (B) = 0.2

P (A ∩ B) = 0.1

Now, we know that,

= P (A) - P (A ∩ B)

= 0.3 – 0.1

= 0.2

Question:41

Answer:

Given:

P (happening of an event A) = 0.5

P (happening of an event B) = 0.3

A & B are mutually exclusive events

Thus, P (A ∩ B) = 0

To find: P (neither A nor B)

= 1 – P (A U B)

= 1 – [P (A) + P (B)]

= 1 – (0.5 + 0.3)

= 1 – 0.8

= 0.2

Question:42

C | C | ||

a | 0.95 | i | An incorrect assignment |

b | 0.02 | ii | no chance of happening |

c | -0.3 | iii | as much chance of happening as not |

d | 0.5 | iv | very likely to happen |

e | 0 | v | very little chance of happening |

Answer:

(a) Since 0.95 is very close to 1, it is very likely to happen.

Thus, (a) → (iv)

(b) The probability of 0.02 is very low, hence there is very little chance of happening, thus, (b) → (v)

(c) Since the probability of -0.3 is negative, and we know that probability can never be negative, thus it is an incorrect assignment, thus, (c) → (i)

(d) For 0.5 the sum of chances of happening and not happening is 1, hence it has as much chance of happening as not, thus, (d) → (iii)

(e) For 0, there is no chance of happening, thus, (e) →(ii)

Question:43

a | if E | i | E |

b | if E | ii | (E |

c | If E | iii | E |

d | If E | iv | E |

Answer:

Answers –

(a) E_{1 }& E_{2} are mutually exclusive events,

Thus, P (E_{1} ∩ E_{2}) = φ

Thus, (a) → (iv)

(b) E_{1} & E_{2} are mutually exclusive and exhaustive events,

Thus, (b) → (iii)

(c) If E_{1} & E_{2} have common outcomes then,

(E_{1} – E_{2}) U (E_{1} ∩ E_{2}) = E_{1}

Thus, (c) → (ii)

(d) If E_{1} & E_{2} are such that,

E_{1} E_{2}

Then, (E_{1} ∩ E_{2}) = E_{1}

Thus, (d) → (i).

NCERT Exemplar Class 11 Maths chapter 16 solutions can be downloaded by the students. They can make use of NCERT Exemplar Class 11 Maths solutions chapter 16 PDF Download function for better convenience. The solutions given in the PDF are explained broadly and every sub-topic is covered.

· Random Experiments

· Outcomes

· Sample space

· Event

· The occurrence of an event

· Types of events

· Algebra of Events

· Mutually Exclusive Events

· Exhaustive Events

· Axiomatic Approach to Probability

· Probability of an Event

· Probability of equally likely outcomes

· Probability of the event ‘A or B’

· Probability of the event ‘not A’

**What will students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability?**

The students can comprehensively understand the different fundamental concepts of Probability like the different types of events and the axiomatic approach used in probability. Pupils who find the formulas of Probability difficult can refer to the NCERT Exemplar Class 11 Maths solutions chapter 16 as all of the formulas along with their application are explained in a stepwise manner.

Illustrations are used which makes the problem even easier to comprehend. MCQ type questions, miscellaneous questions, HOTS, short answer type questions and long answer type questions are included in.

Types of Events in Probability:

To understand the classification of events, the students must first understand what events are. An event is the subset of the sample space. Sure events, complementary events, impossible events are the different types of events that occur in probability. NCERT Exemplar Class 11 Maths solutions Chapter 16 can be used to understand the basic concepts like sample space, subsets and an empty set of probability.

Axiomatic Approach to Probability:

The problems related to mutually exclusive events and exhaustive events are important and should not be skipped by the students. Probability is an important topic in the CBSE mathematics syllabus for Class 11 and has an average weightage. It is one of the most scoring chapters in mathematics and the students can easily mark it just by applying a formula and writing the required steps to solve the problems.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 | Probability |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

**Also Check NCERT Books and NCERT Syllabus here:**

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Download EBook1. Can the solutions of NCERT Exemplar Class 11 Maths Chapter 16 Solutions be downloaded?

Yes, the solutions can be downloaded by using the NCERT Exemplar Class 11 Maths solutions chapter 16 PDF Download function.

2. Can NCERT Exemplar Class 11 Maths Chapter 16 Solutions be refereed for CBSE final exam?

Yes, Class 11 Maths NCERT exemplar solutions chapter 16 can be referred to for CBSE final exam.

3. Which topics should be covered from the chapter Probability?

The following are the important topics of Probability:

· Axiomatic Approach to Probability

· Exhaustive Events

· Mutually Exclusive Events

· Types of Events

· Random Experiments

Sep 06, 2024

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