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#### Question 1:

A box contains 4 white balls and 6 black balls and another one, 3 white and 5 black. We can choose one box and take a ball. If we want a black ball, which box is the better choice?

Number of white balls in the first box = 4

Number of black balls in the first box = 6

Number of white balls in the second box = 3

Number of black balls in the second box = 5

Total number of balls in the first box = 10

Total number of balls in the second box = 8

Probability of getting a black ball from the first box = = Probability of getting a black ball from the second box = = 0.625 > 0.6

Probability of getting a black ball from the first box, i.e., 0.625 > Probability of getting a black ball from the second box, i.e., 0.6

Therefore, the second box is a better choice for getting a black ball.

#### Question 2:

You ask someone to say a (natural) number less than 10. What is the probability that the number is a prime? What is the number asked is to be less than 100?

The numbers from 1 to 9 will be counted as natural numbers less than 10.

There are nine natural numbers less than 10.

The prime numbers from 1 to 9 are 2, 3, 5 and 7.

Total prime numbers = 4

Probability of getting a prime number from the given set of natural numbers

= = Now, if we are asked a number less than 100, then the total natural numbers = 99.

The list of prime numbers less than 100 is:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97

Total number of prime numbers which are less than 100 is 25.

Probability of getting a prime number from the given set of natural numbers

= = #### Question 3:

A box contains paper slips with numbers written on them −4 odd and 5 even. Two more paper slips, one with an odd number and another with an even number are put in. Does the probability of getting an odd number increase of decrease? What about the probability of getting an even number?

Initial number of paper slips which have odd number written on them = 4

Initial number of paper slips which have even number written on them = 5

Probability of getting a slip which has odd number written on it = = Probability of getting a slip which has even number written on it = = Number of paper slips added with odd number written on them = 1

Number of paper slips added with even number written on them = 1

Total number of paper slips which have odd number written on them after addition = 5

Total number of paper slips which have even number written on them after addition = 6

Total number of paper slips in the box after addition of slips = 5 + 6 = 11

Probability of getting a slip which has odd number written on it = Probability of getting a slip which has even number written on it = Initial probability of getting a paper slip which has odd number written on it = Final probability of getting a paper slip which has odd number written on it = Initial probability of getting a paper slip which has even number written on it = Final probability of getting a paper slip which has even number written on it = From the calculated probabilities, we can observe that the probability of getting a paper slip which has odd number written on it has increased from to and the probability of getting a paper slip which has even number written on it has decreased from to .

#### Question 1:

There are two boxes, each containing slips numbered 1 to 5. One slip is drawn from each box and their numbers added. What are the possible sums? Compute the probability of each sum.

Total number of slips in the first box = 5

Total number of slips in the second box = 5

Possible numbers in each box are 1, 2, 3, 4 and 5.

If one slip is drawn from any box, the possible numbers which can be obtained are 1, 2, 3, 4, and 5.

So, the possible sums which can be obtained can be found out by taking all the pairs one by one as follows:

i) When the number drawn from the first box is 1, the pairs obtained are:

 Number Pair Sum (1, 1) 2 (1, 2) 3 (1, 3) 4 (1, 4) 5 (1, 5) 6

ii) When the number drawn from the first box is 2, the pairs obtained are:

 Number Pair Sum (2, 1) 3 (2, 2) 4 (2, 3) 5 (2, 4) 6 (2, 5) 7

iii) When the number drawn from the first box is 3, the pairs obtained are:

 Number Pair Sum (3, 1) 4 (3, 2) 5 (3, 3) 6 (3, 4) 7 (3, 5) 8

iv) When the number drawn from the first box is 4, the pairs obtained are:

 Number Pair Sum (4, 1) 5 (4, 2) 6 (4, 3) 7 (4, 4) 8 (4, 5) 9

v) When the number drawn from the first box is 5, the pairs obtained are:

 Number pair Sum (5, 1) 6 (5, 2) 7 (5, 3) 8 (5, 4) 9 (5, 5) 10

So the sums obtained are 2, 3, 4, 5, 6, 7, 8, 9 and 10.

Probability of obtaining the sum as 2 = Probability of obtaining the sum as 3 = Probability of obtaining the sum as 4 = Probability of obtaining the sum as 5 = Probability of obtaining the sum as 6 = Probability of obtaining the sum as 7 = Probability of obtaining the sum as 8 = Probability of obtaining the sum as 9 = Probability of obtaining the sum as 10 = #### Question 2:

In the finger-raising game, which number has the maximum probability of occurrence as the sum? What is this probability?

In the finger raising game, the sum of number of fingers raised by the opponents is taken into account. For a particular opponent the possibilities of numbers raised range from 1 to 10. Similarly, for the second opponent the possibilities of numbers raised range from 1 to 10.

The pairs can be arranged as:

 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (1, 10) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) (2, 9) (2, 10) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) (3, 9) (3, 10) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8) (4, 9) (4, 10) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8) (5, 9) (5, 10) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8) (6, 9) (6, 10) (7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (7, 7) (7, 8) (7, 9) (7, 10) (8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8) (8, 9) (8, 10) (9, 1) (9, 2) (9, 3) (9, 4) (9, 5) (9, 6) (9, 7) (9, 8) (9, 9) (9, 10) (10, 1) (10, 2) (10, 3) (10, 4) (10, 5) (10, 6) (10, 7) (10, 8) (10, 9) (10,10)

Total number of pairs = 100

The sum can be found out as:

 2 3 4 5 6 7 8 9 10 3 4 5 6 7 8 9 10 12 4 5 6 7 8 9 10 12 13 5 6 7 8 9 10 12 13 14 6 7 8 9 10 12 13 14 15 7 8 9 10 12 13 14 15 16 8 9 10 12 13 14 15 16 17 9 10 12 13 14 15 16 17 18 10 12 13 14 15 16 17 18 19 12 13 14 15 16 17 18 19 20

It can be observed from the pairs that the sum 11 is always involved in every set. So, the number should have the maximum probability.

Probability of occurrence of number 11 = #### Question 3:

Suppose you ask someone to say a two-digit number.

(i)

What is the probability of this number having both digits the same?

(ii)

What is the probability of the first digit being larger than the second?

(iii)

What is the probability of the first digit being smaller than the second?

The numbers ranging from 10 to 99 can be termed as two digit numbers, so there are 90 numbers which have two digits.

(i)

The two digit numbers which have their both the digits as same are 11, 22, 33, 44, 55, 66, 77, 88 and 99.

There are nine numbers of this type.

Probability of obtaining such a number = Probability of obtaining a two digit number which has its both the digits as same = (ii)

The two digit numbers which have their first digit larger than the second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97 and 98.

The total number of such type of numbers is 45.

Probability of obtaining such a number = Probability of obtaining a two digit number which has its first digit larger than the second digit = (iii)

The two digit numbers which have their first digit smaller than the second digit are 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79 and 89.

The total number of such type of numbers is 36.

Probability of obtaining such a number = ∴Probability of obtaining a two digit number which has its first digit smaller than the second digit = View NCERT Solutions for all chapters of Class 10