Mathematics of Chance Class 10 Extra Questions Kerala State Syllabus Maths Chapter 4

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 4 Mathematics of Chance Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 4 Mathematics of Chance Important Questions and Answers

Mathematics of Chance Class 10 Extra Questions Kerala Syllabus

Mathematics of Chance Class 10 Kerala Syllabus Extra Questions

Question 1.
Each letter of the word MALAYALAM are written in small paper pieces and kept in a box. One is taken from the box without looking. The probability of getting A is
(a) \(\frac{2}{9}\)
(b) \(\frac{3}{9}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{1}{9}\)
Answer:
(c) \(\frac{4}{9}\)

Question 2.
The factors of 24 are written in small paper pieces separately and kept in a box. One is taken from the box without looking. The probability of getting even
(a) \(\frac{1}{8}\)
(b) \(\frac{7}{8}\)
(c) \(\frac{6}{8}\)
(d) \(\frac{3}{8}\)
Answer:
(c) \(\frac{6}{8}\)

Question 3.
First 10 terms of the arithmetic sequence 3n + 1 are written in small paper pieces separately and kept in a box.One is taken from the box without looking. What is the probability of getting odd number.
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{5}{11}\)
(d) \(\frac{3}{8}\)
Answer:
(a) \(\frac{1}{2}\)

Question 4.
Radius of the big circle is equal to the diameter of small circle.

A fine dot is placed into the figure without looking The probability of falling the dot in the unshaded part?
Answer:
\(\frac{3}{4}\)

Question 5.
The numbers 2,3,4 are written in small paper pieces kept in one box. Fractions \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) are written in paper pieces and kept in another box. One is taken from each box and find the product
a) What are the outcome pairs
Answer:

Products are 1, \(\frac{2}{3}, \frac{1}{2}, \frac{3}{2}\), 1\(\frac{3}{4}\), 2, \(\frac{4}{3}\), 1

b) What is the probability of getting an integer as the product?
Answer:
\(\frac{4}{9}\)

Question 6.
Numbers 1, 2, 3,…25 are written in small paper pieces and put in a box.One is taken from it without looking into the box.
a) What is the probability of getting an even number?
Answer:
\(\frac{12}{25}\)

b) What is the probability of getting an odd number?
Answer:
\(\frac{13}{25}\)

c) What is the probability of getting a prime number?
Answer:
\(\frac{9}{25}\)

Question 7.
A die in which the numbers 1 to 6 are written on the faces is thrown
a) What is the probability of falling an even numbered face?
Answer:
Probability of falling even face = \(\frac{3}{6}=\frac{1}{2}\)

b) What is the probability of getting an odd numbered face?
Answer:
Probability of falling odd face = \(\frac{3}{6}=\frac{1}{2}\)

c) What is the probability of getting a prime numbered face?
Answer:
Probability of falling prime numbered face = \(\frac{3}{6}=\frac{1}{2}\)

Question 8.
Two digit numbers are written in small paper pieces and placed in a box. One is taken from the box at random
a) How many multiples of 5 are there in the box?
Answer:
10, 11, 12, ……….. 99 are the two digit numbers.
Number of two digit numbers is 90
Multiples of five are 10,15, 20…95
Number of numbers = 18

b) What is the probability of getting a multiple of 5?
Answer:
Probability of getting a multiple of five = \(\frac{18}{90}\)

c) What is the probability of not getting a multiple of 5 ?
Answer:
Probability of not getting a multiple of 5
= 1 – \(\frac{18}{90}=\frac{72}{90}=\frac{8}{10}\)

Question 9.
Two dice numbered 1 to 6 are thrown at together.
a) Write the outcomes as pairs
Answer:
(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)

b) What is the probability of the occurence of equal numbers?
Answer:
\(\frac{6}{36}=\frac{1}{6}\)

c) What is the probability of the occurence of perfect squares ?
Answer:
(1, 1), (1, 4), (4, 1), (4, 4).
Probability \(\frac{4}{36}\)

d) What is the probability of the occurence of multiple of 2 in one die and multiple of 3 in other die ?
Answer:
(2, 3), (4, 3), (6, 3), (2, 6), (4, 6), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)
Probability \(\frac{11}{36}\)

Question 10.
What is the probability of getting 5 Sundays in the month December?
Answer:
There are 31 days in December. 28 days decide 4weeks, so four Mondays.
The combinations are (Sunday, Monday, Tuesday), (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thusday, Friday), (Thusday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday).
There are three combinations in which Sundays occur.Probability of occuring five Sundays is \(\frac{3}{7}\)

Question 11.
The numbers 2¹, 2², 2³… 2⁵⁰ are written in small paper pieces and placed in a box.
a) Write the sequence of numbers comes in the right end of these numbers ?
Answer:
2, 4, 8, 6, 2, 4, 8, 6…

b) If one is taken from the box at random, then what is the probability of getting a number with 4 in ones place?
Answer:
Up to 2⁴⁸, the set of digits 2, 4, 8, 6 repeats 12 times. One’s place of 2⁴⁹ is 2 and one’s place of 2⁵⁰ is 4
Probability of getting 4 in one’s place is = \(\frac{13}{50}\)

c) What is the probability of getting a number with 8 in ones place?
Answer:
Probability of getting 8 in one’s place is = \(\frac{12}{50}\)

d) What is the probability of getting a number with 2 in ones place?
Answer:
Probability of getting 2 in one’s place is = \(\frac{13}{50}\)

e) What is the probability of not getting a number with 2 in ones place?
Answer:
Probability of not getting 2 in one’s place is = 1 – \(\frac{13}{50}\)
= \(\frac{37}{50}\)

Question 12.
Two digit numbers are written in small paper pieces and placed in a box.
a) How many paper slips are there in the box?
b) If one is taken from the box, what is probability of getting a number with digits same?
c) If one is taken from the box, what is probability of getting a number in which the product of the digits a prime number.
d) What is the probability of getting a prime number?
Answer:
a) 10, 11, 12…99 are the numbers. There are 90 numbers

b) Numbers with same digits are 11, 22, 33, 44, 55, 66, 77, 88, 99
Total number of these numbers is 9 Probability = \(\frac{9}{90}\) = \(\frac{1}{10}\)

b) In the two digit numbers with product of the digits a prime, one digit is 1 and other digit is one of the numbers 2, 3, 5, 7

c) Numbers are 12, 13, 15, 17, 21,31,51,71.
Probability = \(\frac{8}{90}=\frac{4}{45}\)

d) There are 25 prime numbers below 100.4 of them are one digit primes and the rest of the 21 numbers are two digit primes.
Probability is = \(\frac{21}{90}=\frac{7}{30}\)

Question 13.
There are two circles in the picture.One is inside other.Radius of the small circle is half of the radius of the big circle.

a) If the radius of the small circle is r then what is the area of the small circle and big circle ?
b) If a fine dot is placed into the figure, what is the probability of falling the dot in the small circle?
c) What is the probability of falling the dot the
yellow shaded part in the figure.
Answer:
a) Area of small circle πr²
Area of big circle π × (2r)² = 4πr²
b) Chance of getting black \(\frac{\pi r^2}{4 \pi r^2}=\frac{1}{4}\)
c) Probability of falling the dot in the yellow shade is 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 15.
Triangle PQR is drawn by joining the midpoints of the sides of triangle ABC.

a) How many equal triangles are there in the figure?
Answer:
4 triangles.
∆PQR, ∆APQ, ∆PCR, ∆QRB are equal triangles

b) A fine dot is placed into the figure. What is the probability of falling the dot in triangle PQR?
Answer:
\(\frac{1}{4}\)

c) How many parallelograms are there in the picture?
Answer:
3 Parallelograms
PQRC, PQBR, PRQA are equal

d) A fine dot is placed into the figure. What is the probability of falling the dot in the parallelogram PQRC ?
Answer:
To fall the dot in PQRC it is necessary to fall either in triangle PCR or triangle PQR
Probability is \(\frac{2}{4}=\frac{1}{2}\)

Question 15.
ACP is drawn in the square ABCD and shaded. P is the mid point of the side of the square

a) If the side of the square is ‘a’ then what is the altitude to the side PC of the shaded triangle.
Answer:
AB = a

b) If the side of the square is V then what is the area of the shaded triangle?
Answer:
Height of triangle APC is = \(\frac{a}{2}\)
Height = a
Area = \(\frac{1}{2} \times \frac{a}{2}\) × a = \(\frac{\mathrm{a}^2}{4}\)

c) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?
Answer:
Probability = \(\frac{\mathrm{a}^2}{4}\) ÷ a = \(\frac{1}{4}\)

Question 16.
There are two squares in the figure.The perimetre of the outer square is 28 cm , the perimetre of the inner square is 20 cm

a) What is the area of the outer square?
Answer:
Side of outer square = \(\frac{28}{4}\) = 7 cm
Area = 7² = 49 sq.cm

b) What is the area of inner square?
Answer:
Side of inner square \(\frac{20}{4}\) = 5 cm
Area = 5² = 25 sq.cm

c) What is the area of the shaded triangle ?
Answer:
Area of the region in between the squares
= 49 – 25 = 24 sq.cm
Area of shaded part = \(\frac{24}{4}\) = 6 sq.cm

d) If a fine dot is placed into the figure then what is the probability of falling the dot in the shaded triangle?
Answer:
Probability = \(\frac{6}{49}\)

Question 17.
The mid points of the two sides and one vertex of a square are joined in such a way as to get a triangle which is coloured in the picture.

a) If the side of the square is a, what is area of the unshaded triangles ?
Answer:
Unshaded part = (\(\frac{1}{2}\) × a × \(\frac{a}{2}\)) × 2 + \(\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2}\)
= \(\frac{a^2}{2}+\frac{a^2}{8}=\frac{5 a^2}{8}\)

b) What is the area of the shaded triangle?
Answer:
Area of shaded part = a – \(\frac{5 a^2}{8}=\frac{3 a^2}{8}\)

c) If a fine dot is placed into the figure then what is the probability of falling the dot in the coloured traingle?
Answer:
Probability = \(\frac{3 a^2}{8}\) ÷ a² = \(\frac{3}{8}\)

Question 18.
Manju has three ornaments :Green, Red and Blue ear rings and chains.She ware it in different ways.
a) How many ways she can ware the ornaments?
b) What is the probability of waring ornaments of same colour?
c) What is the probability of wearing the orna-ments of different colours?
Answer:
a) Number of pairs 3 × 3 = 9, (Green, Green), (Green , Red), (Green, Blus) (Blue, Green), (Blue, Red), (Blue, Blue) (Red, Green), (Red, Red), (Red, Blue)
b) (Green, Green), (Red, Red), (Blue, Blue) Probability = \(\frac{3}{9}=\frac{1}{3}\)
c) Probability of wearing different colours is 1 – \(\frac{1}{3}=\frac{2}{3}\)

Question 19.
A box contains 4 black balls and 3 white balls. Another box contains 5 black balls and 3 white balls. One from each box is taken at random.
a) How many pair of balls are possible ?
b) What is the probability of getting both balls black?
c) What is the probability of getting both balls white?
d) What is the probability of getting balls of dif-ferent colours?
Answer:
a) Total number of possible selections = (3 + 4) × (5 + 3) = 7 × 8 = 56

b) Probability of getting both black
\(\frac{4 \times 5}{56}=\frac{20}{56}=\frac{5}{14}\)

c) Probability of getting both white \(\frac{3 \times 3}{56}=\frac{9}{56}\)

d) Probability of getting balls of different colours \(\frac{(4 \times 3)+(3 \times 5)}{56}=\frac{27}{56}\)

Question 20.
A box contains four paper slips carrying numbers 1,2,3,4. Another box contains paper slips carrying numbers 1, 2, 3. One from each box is taken at random and entered as pairs.
a) How many pairs are possible ?
b) What is the probability of getting a pair with the product of the digits odd?
c) What is the probability of getting a pair with the product of the digits even?
Answer:
a) Number of pairs 4 × 3 = 12
(1, 1) , (1, 2), (1, 3)
(2, 1), (2, 2), (2, 3)
(3, 1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)

b) Pairs of getting a product odd are
(1, 1), (1, 3), (3, 1), (3, 3)
Probability \(\frac{4}{12}=\frac{1}{3}\)

c) Probability of getting product even = 1 – \(\frac{1}{3}=\frac{2}{3}\)

Question 21.
A bag contains 4 black beads and 3 white beads.Another bag contains 4 black beads and 5 white beads. One is taken from each bag
a) What is the probability of getting both white ?
b) What is the probability of getting both black ?
c) What is the probability of getting one white and one black?
d) What is the probability of getting atleast one black?
Answer:
a) First bag contains 7 beads and second bag con-tains 9 beads. When one from each box are taken we get 7×9 = 63 pairs.
Number of pairs with both white is, 3 × 5 = 15
Probability of getting both white is \(\frac{15}{63}=\frac{5}{21}\)

b) Number of pairs with both black is, 4 × 4 = 16
Probability of getting both black is \(\frac{16}{63}\)

c) Probability of getting one white and one black is \(\frac{4 \times 5+3 \times 4}{63}=\frac{32}{63}\)

d) Probability of getting atlest one black is
\(\frac{4 \times 5+3 \times 4+4 \times 4}{63}=\frac{48}{63}=\frac{16}{21}\)

Question 22.
A box contains four paper strips on which the numbers 1,2,3,4 are written. Another box contains the strips 1, 2, 3. One is taken from each box
a) What are the possible outcomes
Answer:
(1, 1), (1, 2), (1, 3)
(2, 1), (2, 2), (2, 3)
(3, 1), (3, 2), (3, 3)
(4, 1), (4, 2), (4, 3)

b) What is the probability of getting the sum a multiple of 3 ?
Answer:
Favourable outcomes are
(1, 2) , (2, 1), (3, 3), (4, 2)
Probability is = \(\frac{4}{12}=\frac{1}{3}\)

c) What is the probability of getting the sum a multiple of 2 ?
Answer:
Favourable outcomes are
(1, 1), (3, 1), (1, 3), (2, 2), (3, 3), (4, 2)
Probability is \(\frac{6}{12}=\frac{1}{2}\)

d) What is the probability of getting the product a multiple of 6
Answer:
Favourable outcomes are (2,3), (3,2), (4,3)
Probability is \(\frac{3}{12}=\frac{1}{4}\)