# RD SHARMA Solutions for Class 10 Maths Chapter 16 - Probability

## Chapter 16 - Probability Exercise Ex. 16.1

A coin has only two options-head and tail and both are equally likely events i.e. the probability of occurrence of both is same. Hence, a coin is a fair option to decide which team will choose ends in the game.

(i) 1

(ii) 0

(iii) 0 and 1

(iv) equal

(v) 1

(vi) 1

(x)

When a black die and a white die are thrown at the same time, the sample space is given by

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

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

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

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

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

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

n(S) = 36

Let A be the event that the numbers obtained have a product less than 16.

A = {{(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),

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

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

n(A) = 25

P(A) =

Sample space when three coins are tossed together is

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

n(S) = 8

Let A be the event ofgetting exactly two heads.

A = {HHT, THH, HHT}

n (A) = 3

P(A) =

Sample space when three coins are tossed together is

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

n(S) = 8

Let A be the event of getting at most two heads.

A = {TTT, HTT, THT, TTH, HHT, THH, HHT}

n (A) = 7

P(A) =

Sample space when three coins are tossed together is

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

n(S) = 8

Let A be the event of getting at least one head and one tail.

A = {HHT, HTH, THH, HTT, THT, TTH }

n (A) = 6

P(A) =

Sample space when three coins are tossed together is

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

n(S) = 8

Let A be the event ofgetting no tails.

A = {HHH}

n (A) = 1

P(A) =

## Chapter 16 - Probability Exercise Ex. 16.2

Assume first circle to be the circle with the smallest radius, that is 3. Similarly, second circle to be the circle with radius 7 and third circle to be the circle with radius 9.

## Chapter 16 - Probability Exercise 16.35

n(E) = total numbers

= 9

n(0) = odd numbers {1, 3, 5, 7, 9}

= 5

So, the correct option is (b).

n(E) = 9

n(4) = no. is even {2, 4, 6, 8}

= 4

So, the correct option is (a).

## Chapter 16 - Probability Exercise 16.36

n(E) = 9

n(A) = no. is multiple of 3 {3, 6, 9}

= 3

So, the correct option is (a).

3 coins are tossed simultaneously.

Hence sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Event (E) = at least two Heads

= {HHH, HHT, HTH, THH}

n(s) = 8

n(E) = 4

So, the correct option is (c).

sample space (s) = {1, 2, 3, 4, 5, 6}

n(s) = 6

Event (E) = getting a multiple of 3

= {3, 6}

n(E) = 2

So, the correct option is (b).

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

n(S) = 6

Event (E) = getting number greater than 2

= {3, 4, 5, 6}

n(E) = 4

So, the correct option is (c).

n(S) = 52

no. of ace in a pack of 52 cards = 4

n(E) = 4

So, the correct option is (b).

n(S) = 25

Event (E) = prime numbers between 1 to 25

= {2, 3, 5, 7, 11, 13, 17, 19, 23}

n(E) = 9

Note: The answer does not match the options in the question.

We know probability P(E) of an event lies between 0 < P(E) < 1 ......(1)

(a), (c), (d) satisfies the (1) but (b) is a negative number. It can't be the probability of an event.

So, the correct option is (b).

We know

P(E) + P(not E) = 1

given P(E) = 0.05

so P(not E) = 1 - 0.05

= 0.95

So, the correct option is (d).

We know 0 < P(E) < 1

(a), (b), (c) fullfill the condition. But (d) doesn't

Hence (d) is correct option.

So, the correct option is (d).

An event that is certain to occur is called Certain event.

Probability of certain event is 1.

Ex: If it is Monday, the probability that tomorrow is Tuesday is certain and therefore probability is 1.

So, the correct option is (b).

## Chapter 16 - Probability Exercise 16.37

Events that are not possible are impossible event.

Probability of impossible event is 0.

So, the correct option is (a).

## Chapter 16 - Probability Exercise 16.38

Sample space (s) = {-3, -2, -1, 0, 1, 2, 3}

n(s) = 7

Event (E) = |x| < 2

= {-1, 0, 1}

n(E) = 3

So, the correct option is (c).

There are 365 days in a non-leap year.

52 complete weeks and 1 spare day.

so This day can be any out of 7 day of week.

Hence n(s) = 7

Now, year already have 52 Sundays. so for a total of 53 Sundays in a calendar year, this spare day must be a Sunday.

Hence n(E) = 1

So, the correct option is (d).

We know on throwing a pair of die there are a total of 36 possible outcomes.

n(S) = sum is a perfect square

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

n(E) = 7

So, the correct option is (b).

There are 365 days in a non-leap year.

52 complete weeks and 1 spare day.

So this day can be any out of 7 day of a week.

Hence n(s) = 7

Now, a non-leap year already has 52 Sundays. So for a total of 53 Sundays in a calendar year, this spare day must be a Sunday.

Hence n(E) = 1

So, the correct option is (b).

## Chapter 16 - Probability Exercise 16.39

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics### RD SHARMA Solutions for CBSE Class 10 Subjects

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