Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 4 Mathematics of Chance Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 4 Mathematics of Chance Questions and Answers
Mathematics of Chance Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 4 Mathematics of Chance Solutions
Class 10 Maths Chapter 4 Kerala Syllabus – Chances As Numbers
(Textbook Page No. 74, 75)
Question 1.
A box contains 6 black and 4 white balls. If a ball is picked from it, what is the probability that it is black? And the probability that it is white?
Answer:
(a) \(\frac{6}{10}=\frac{3}{5}\)
(b) \(\frac{4}{10}=\frac{2}{5}\)
Question 2.
A bag contains 3 red balls and 7 green balls. Another contains 8 red and 7 green balls
(a) If a ball is drawn from the first bag what is the probability that it is red?
Answer:
\(\frac{3}{10}\)
(b) From the second bag?
Answer:
\(\frac{8}{15}\)
(c) The balls in both bags are put together in a single bag. If a ball is drawn from this, what is the probability that it is red?
Answer:
\(\frac{11}{25}\)
Question 3.
A bag contains 3 red beads and 7 green beads. Another bag contains one more of each. The probability of getting a red from which bag is greater?
Answer:
From the first bag, probability of getting red is \(\frac{3}{10}\)
In second bag, Number of red balls 4 , number of green balls 8
Probability of getting red \(\frac{4}{12}=\frac{2}{6}\)
\(\frac{3}{10}=\frac{18}{60}\), \(\frac{2}{6}=\frac{20}{60}\)
\(\frac{20}{60}>\frac{18}{60}\)
Taking ball from the second bag is more probable to get red
Class 10 Maths Kerala Syllabus Chapter 4 Solutions – Number Probability
(Textbook Page No. 76)
Question 1.
In each of the problems below, compute the probability as a fraction and then write it in decimal form and as a percent
(i) If one number from 1, 2, 3, …, 10 is chosen, what is the probability that it is a prime number?
(ii) If one number from 1, 2, 3, …, 100 is chosen, what is the probability that it is a two-digit number?
(iii) Every three digit number is written in a slip of paper, and all the slips are put in a box. If one slip is drawn, what is the probability that it is a palindrome?
Answer:
(i) \(\frac{4}{10}\) 0.4 40%
(ii) \(\frac{90}{100}=\frac{9}{10}\), 0.9,90%\frac{90}{100}=\frac{9}{10}
(iii) \(\frac{10}{100}=\frac{1}{10}\), 0.1, 10%
Question 2.
A person is asked to say a two-digit number. What is the probability that is a perfect square?
Answer:
\(\frac{6}{90}=\frac{1}{15}\)
Question 3.
A person is asked to say a three-digit number.
(i) What is the probability that all three digits of this number are the same?
Answer:
\(\frac{9}{900}=\frac{1}{100}\)
(ii) What is the probability that the digit in one’s place of this number is zero?
Answer:
\(\frac{90}{900}=\frac{1}{10}\)
(iii) What is the probability that this number is a multiple of 3?
Answer:
\(\frac{300}{900}=\frac{1}{3}\)
Question 4.
Four cards with numbers 1, 2, 3, 4 on them, are joined to make a four-digit number.
(i) What is the probability that the number is greater than four thousand?
Answer:
There are 24 such numbers. 6 of them are more than 4000 Probability of getting more than 4000 is \(\frac{6}{24}=\frac{1}{4}\)
(ii) What is the probability that the number is less than four thousand?
Answer:
\(\frac{18}{24}=\frac{3}{4}\)
SCERT Class 10 Maths Chapter 4 Solutions – Geometrical Probability
(Textbook Page No. 78)
In each picture below, the description of the green region inside the yellow one is given. In each, find the probability that a dot put in the picture, without looking, falls within the green region
Question 1.
The square got by joining the midpoints of a larger square
Answer:
The square got by joining the midpoints of a larger square
Draw horizontal and vertical lines. It divides the figure into 8 equal parts. Four of them are shaded.So probability is \(\frac{1}{2}\).
Question 2.
The triangle got by joining alternate vertices of a regular hexagon
Answer:
The triangle got by joining alternate vertices of a regular hexagon
Divide the regular hexagon into six equal triangles.
Three of them are green probability is \(\frac{3}{6}=\frac{1}{2}\)
Question 3.
The regular hexagon formed between two equal equilateral triangle.
Answer:
Regular hexagons formed by two equilateral triangles
Divide the green hexagon into six equal triangles.
These are 6 among 12 equal triangles.Probability is \(\frac{6}{12}=\frac{1}{2}\)
Question 4.
A square drawn with vertices on a circle
Answer:
A square drawn with vertices on the circle Let a be the side of the square. Diagonal √2 a
Radius of the circle \(\frac{a}{\sqrt{2}}\)
Area of the circle π\(\frac{a^2}{2}\)
Probability is \(\frac{a^2}{\frac{\pi a^2}{2}}=\frac{2}{\pi}\)
Question 5.
The circle that just fits within a square
Answer:
The circle that just fits within a square
Radius of the circle r.
Area = πr2
Side of the square 2r. Area of the square 4r2;
Probability = \(\frac{\pi}{4}\)
SSLC Maths Chapter 4 Questions and Answers – Pairs
(Textbook Page No. 80, 81)
Question 1.
Rajani has three necklaces and three earrings of green, blue and red stones. In how many different ways can she wear them ? What is the probability of her wearing a necklace and earrings of the same colour ? Of different colours?
Answer:
Different ways: 3 × 3 = 9
Probability of wearing a necklace and carring of same colour = \(\frac{3}{9}=\frac{1}{3}\)
Probability of wearing a necklace and carring of different colour = \(\frac{6}{9}=\frac{2}{3}\)
Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1 and 2. If one slip is drawn from each box, what is the probability of the sum of the numbers being odd? What is the probability of the sum being even?
Answer:
(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2) Pairs having sum odd are (1, 2), (2, 1), (3, 2) (4, 1)
Probability of getting sum odd = \(\frac{4}{8}=\frac{1}{2}\)
Probability of getting sum even = \(\frac{4}{8}=\frac{1}{2}\)
Question 3.
A box contains four slips numbered 1, 2, 3, 4, and another box contains three slips numbered 1, 2, 3. If one slip is drawn from each box, what is the probability of the product of the numbers being odd? What is the probability of the product being even?
Answer:
Possible outcomes 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)
Probability of getting product odd = \(\frac{1}{3}\)
Probability of getting product even = \(\frac{2}{3}\)
Question 4.
From all two-digit numbers using only the digits 1, 2, and 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?
Answer:
(i) Outcomes are 11, 12, 13, 21, 22, 23, 31, 32, 33
Probability is \(\frac{3}{9}=\frac{1}{3}\)
(ii) \(\frac{3}{9}=\frac{1}{3}\)
Question 5.
A game for two players. Before starting, each player has to decide whether he wants an odd number or even number. Then both raise some fingers of one hand at the same time. If the sum of numbers of fingers is odd, the one who chose odd number at the beginning wins; if even, the one who chose even number wins. Which is the better choice at the beginning, odd or even?
Answer:
Pairs 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)
Sums 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
Probability of getting the sum even = \(\frac{13}{25}\)
Probability of getting the sum odd = \(\frac{12}{25}\)
Choosing even at the beginning is better.
Kerala Syllabus Class 10 Maths Chapter 4 Solutions – More Pairs
(Textbook Page No. 84)
Question 1.
There are 30 boys and 20 girls in Class 10 A and 15 boys and 25 girls in Class 10 B. One student is to be chosen from each Class.
(i) What is the probability of both being girls ?
Answer:
There are 30 boys and 20 girls in Class 10 A and 15 boys and 25 girls in Class 10 B. One student is to be chosen from each Class.
\(\frac{20 \times 25}{2000}=\frac{500}{2000}=\frac{1}{4}\)
(ii) What is the probability of both being boys ?
Answer:
\(\frac{30 \times 15}{2000}=\frac{450}{2000}=\frac{9}{40}\)
(iii) What is the probability of one being a boy and one being a girl?
Answer:
\(\frac{30 \times 25+20 \times 15}{2000}=\frac{105}{200}=\frac{21}{40}\)
(iv) What is the probability of at least one being a boy?
Answer:
\(\frac{1500}{2000}=\frac{15}{20}=\frac{3}{4}\)
Question 2.
One is asked to say a two-digit number
(i) What is the probability of both digits being the same ?
Answer:
\(\frac{9}{90}=\frac{1}{10}\)
(ii) What is the probability of the first digit being greater than the second?
Answer:
Without actual listing
10, (20, 21), (30, 31, 32)…. (90, 91, 92….98)
Total numbers = 1 + 2 + 3 + …. + 9 = 45
Probability is \(\frac{45}{90}\)
(iii) What is the probability of the first digit being less than the second?
Answer:
8 + 7 + 6 + 5 + 4 + 3 + 2+1=36
Probability is \(\frac{36}{90}\)
Question 3.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums that can be got ? Which sum has the maximum possibility?
Answer:
Sums are listed below
2, 3, 4, 5, 6, 7
3, 4, 5, 6, 7, 8
4, 5, 6, 7, 8, 9
5, 6, 7, 8, 9, 10
6, 7, 8, 9, 10, 11
7, 8, 9, 10, 11, 12
Question 4.
One box contains 10 slips numbered 1 to 10 and another contains slips numbered with multiples of 5, up to 25. One slip is drawn from each box.
(i) What is the probability of both numbers being odd?
(ii) What is the probability of the product of the numbers being odd ?
(iii) What is the probability of the sum of the numbers being odd?
Answer:
(i) 5 odd numbers in the first box and 3 odds in the second box. Probability is \(\frac{15}{50}=\frac{3}{10}\)
(ii) To get product odd, number from first box is odd and number from second box odd
Probability is \(\frac{15}{50}=\frac{3}{10}\)
(iii) To get sum odd, numbers from the first box odd from the second box even or number from the first box even and number from second box is odd.
Probability is \(\frac{5 \times 2+5 \times 3}{50}=\frac{25}{50}=\frac{1}{2}\)
Mathematics of Chance Class 10 Notes Pdf
Class 10 Maths Chapter 4 Mathematics of Chance Notes Kerala Syllabus
Introduction
The unit ‘Mathematics of Chance’ or ‘Probability’ discusses the numerical measurement of chance. There are experiments whose outputs cannot be predicted. Such experiments are generally known as probability experiments.
Tossing a coin and throwing a dice are probability experiments. The outcomes in the tossing a coin are Head H and Tail T. We define probability of getting H on a single toss as \(\frac{1}{2}\) .
There is a section ‘Geometric probability’ in the unit. It measures probability as the ratio of areas.
The third section of the unit is related to fundamental principle of counting. If an activity can be performed in m ways, another activity by n ways then both the activities can be performed in m × n ways. Questions related to this situation is a part of the unit.
→ The experiments whose results cannot be predicted are called probability experiments. Tossing a coin, throwing a dice are the examples of probability experiments
→ On tossing a coin the expected outcomes are Head and Tail. The probability of getting head on tossing is \(\frac{1}{2}\)
→ Throwing a dice is another experiment. Dice is a cubical object numbered 1 to 6 on its faces. The expected outcomes are 1, 2, 3, 4, 5, 6. On throwing, the probability of getting a prime numbered face is \(\frac{3}{6}\)
→ Probability is the measure of chance. Sometimes by comparing area of geometric shapes also comes as the part of probability experiments.
For example, we know that the diagonal of a rectangle divides the rectangle into two equal triangles.Let one of them be shaded. A fine dot is placed into the rectangle without looking in it. The probability of falling the dot in the shaded part is the ratio of area of shaded part to the area of the rectangle. It is \(\frac{1}{2}\).
→ There are situations to measure probability by pairing of objects.
A box contains 3 black balls and 3 white balls. Another box contains 4 black balls and 2 white balls One ball is taken from each box and write as pairs
Number of pairs is (3 + 3) × (4 + 2) = 6 × 6 = 36
Probability of getting both white = \(\frac{3 \times 2}{36}\)
Probability of getting both black = \(\frac{3 \times 4}{36}\)
Probability of getting one black and one white = \(\frac{3 \times 2 + 3 \times 4}{36}\)
→ Probability Experiments
- Experiments where results cannot be predicted in advance are called probability experiments.
- Examples: Tossing a coin, throwing a dice.
→ Tossing a Coin
- Possible outcomes: Head (H) or Tail (T).
- Probability of getting a Head = \(\frac{1}{2}\), and Tail = \(\frac{1}{2}\)
→ Throwing a Dice
- A standard dice has six faces numbered from 1 to 6
- Possible outcomes: 1, 2, 3, 4, 5, 6.
- Prime numbers among them: 2, 3, 5