# Class 10 RD SHARMA Solutions Maths Chapter 16 - Probability

## Probability Exercise Ex. 16.1

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### Solution 6(xix)

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### Solution 19(i)

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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.

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(i) 1

(ii) 0

(iii) 0 and 1

(iv) equal

(v) 1

(vi) 1

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### Solution 50(xvi)

### Solution 50(xvii)

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### Solution 53

(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) =

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### Solution 3(i)

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) =

### Solution 3(ii)

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) =

### Solution 3(iii)

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) =

### Solution 3(iv)

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) =

## Probability Exercise Ex. 16.2

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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.

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## Probability Exercise 16.35

### Solution 1

n(E) = total numbers

= 9

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

= 5

So, the correct option is (b).

### Solution 2

n(E) = 9

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

= 4

So, the correct option is (a).

## Probability Exercise 16.36

### Solution 3

n(E) = 9

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

= 3

So, the correct option is (a).

### Solution 4

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).

### Solution 5

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).

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### Solution 9

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).

### Solution 10

n(S) = 52

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

n(E) = 4

So, the correct option is (b).

### Solution 11

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.

### Solution 12

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).

### Solution 13

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).

### Solution 14

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).

### Solution 15

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).

## Probability Exercise 16.37

### Solution 16

Events that are not possible are impossible event.

Probability of impossible event is 0.

So, the correct option is (a).

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## Probability Exercise 16.38

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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).

### Solution 30

### Solution 31

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).

### Solution 32

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).

### Solution 33

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).

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## Probability Exercise 16.39

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