SCERT Kerala Class 10 Mathematics/ Mathematics of Chance.

SCERT Kerala Class 10 Mathematics / Mathematics of Chance is about the textbook solutions of Class 10 Mathematics. Here you can find out practice problems for the chapter Mathematics of Chance.

SCERT Kerala Class 10 Mathematics Textbook Solutions

Mathematics of Chance – Chapter 3
Answer the following:

1. Rajani has three necklaces and three pairs of earrings, of green, blue and red stones. In what all different ways can she wear them? What is the probability of her wearing the necklace and earrings of the same colour? Of different colours?
   Solution:
   Possible ways of wearing 3 necklaces and 3 pairs of earrings of green, blue and red stones are:
   Necklace: Green
   Earrings: Green, Blue, Red
   Necklace: Blue
   Earrings: Green, Blue, Red
   Necklace: Red
   Earrings: Green, Blue, Red
   These are the 9 different ways.
   Probability of wearing the necklace and earrings of the same colousr = 3/9 = 1/3
   Probability of wearing the necklaces and earrings of different colours = 6/9 = 2/3
2. A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1, 2. If one slip is taken from each, what is the probability of the sum of numbers being odd? What is the probability of the sum being even?
   Solution:
   First box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1, 2.
   If one slip is taken from each, the possible ways are (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)
   Sum of these pairs are 2, 3, 3, 4, 4, 5, 5, 6
   Out of these 8 sums, Probability of the sum of numbers being odd = 4/8 = ½
   Probability of the sum of numbers being even = 4/8 = ½
3. A box contains four slips numbered 1, 2, 3 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even?
   Solution:
   First box contains four slips numbered 1, 2, 3, 4 and another box contains three slips numbered 1, 2, 3.
   If one slip is taken from each, the possible ways are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2) , (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)
   Products of these pairs are 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, 12.
   Out of these 12 products, Probability of the product being odd = 4/12 = 1/3
   Probability of the product being even = 8/12 = 2/3
4. From all two-digit numbers with either digit 1, 2 or 3 one number is chosen.
   i) What is the probability of both digits being the same?
   ii) What is the probability of the sum of the digits being 4?
   Solution:
   Possible two-digit numbers are 11, 12, 13, 21, 22, 23, 31, 32, 33
   Out of these 9 ways, Probability of both digits being the same = 3/9 = 1/3
   Probability of the sum of the digits being 4 = 3/9 = 1/3
5. A game for two players. First, each has to decide whether he wants odd number or even number. Then both raises some fingers of one hand. If the sum is odd, the one who chose odd at the beginning wins: if it is even, the one who chose even wins. In this game, which is the better choice at the beginning, odd or even?
   Solution:
   Possible ways are;
   (1, 1), (1, 2), (1, 3), (1, 4), (1, 5)
   (2, 1), (2, 2), (2, 3), (2, 4), (2, 5)
   (3, 1), (3, 2), (3, 3), (3, 4), (3, 5)
   (4, 1), (4, 2), (4, 3), (4, 4), (4, 5)
   (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)
   Sum of the pairs are
   2, 3, 4, 5, 6
   3, 4, 5, 6, 7
   4, 5, 6, 7, 8
   5, 6, 7, 8, 9
   6, 7, 8, 9, 10
   Among these 25 sums, the number of even numbers are 13 and number of odd numbers are 12.Therefore, Probability of the sum being even = 13/25
   Probability of the sum being odd = 12/25
   So even is the better choice.