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Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:00 PM IST

**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

Mar 02, 2024

Mar 02, 2024

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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

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Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns.

2 Jobs Available

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

3 Jobs Available

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available

2 Jobs Available

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available

5 Jobs Available

3 Jobs Available

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

The procurement Manager is also known as Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

3 Jobs Available

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available

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