Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.3 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 8 – Statistics and Probability
Board  TNSCERT Class 10th Maths 
Class  10 
Subject  Maths 
Chapter  8 (Exercise 8.3) 
Chapter Name  Statistics and Probability 
TNSCERT Class 10th Maths Pdf  all Exercise Solution
Exercise – 8.3
(1) Write the sample space for tossing three coins using tree diagram.
Solution:
We know that each coin tossing can tain two focus (Head and Tail)
Hence the samples space can be written as
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(2) Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Solution:
Given that, two balls in a bag and balls containing numbers 1 to 6
Hence, the sample space can be written as
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)}
(3) If A is an event of a random experiment such that P (A) : P(A) = 17:15 and n (S) = 640 then find
(i) P(A)
(ii) n (A).
Solution:
Given, that P(A):P(A) = 17:15
We know that P(A) + P(A) = 2
P (A) = 1 – P(A) —– (i)
Now, P(A)/P(A) = 17/15
1 – P(A)/P(A) = 17/15 [By equation (i)]
15 – 15P(A) = 17P(A)
P(A) (15 + 17) = 15
P (A) = 15/32
Thus, the P(A) = 15/32
(ii) Now, P(A) = 1 – P (A) = 1 – 15/32 = 17/32
Now, P (A) = n(A)/n(S) = 17/32
Given that, N (S) = 640
∴P(A) = n(A)/n(s) = 17/32
n(A)/640 = 17/32
∴n(A) = 17×640/32
n(A) = 17×20 = 340
Thus, n(A) = 340
(4) A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
A coin is tossed thrice sample Space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, PTT}
∴n (S) = 8
Let, the event of getting two consecutive tails be A.
A = {HTT, TTH, TTT}
n (A) = 3
P (A) = n(A)/n(S) = 3/8 [∵ P(A) = n(A)/n(S)]
Thus, 3/8 probability of getting two consecutive tails.
(5) At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that (i) the first player wins a prize (ii) the second player wins a prize, if the first has won?
Solution:
We know that 23 to 31 squaring contain from 500 to 1000.
(i) Then, n(A) = 9
[Let A be event first player wins a prize]
∴n (S) = 1000
P (A) = n(A)/n(S) = 9/1000
P (First player wins a prize) = 9/1000
(ii) Let B be the event the second player wins a prize if the first has won.
Now, first has won
So, one card is already selected from 500 to 1000 given that repetition is not allowed.
Then one card is remove.
∴n (S) = 1000
∴n (S_{1}) = 1000 – 1 = 999
and n (B) = 9 – 1 = 8
P (B) = n(B)/n(S_{1}) = 8/999
P (second player wins a prize if the first has won) = 8/999
(6) A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Solution:
(i) Blue balls = 12 and red balls = x
∴ Sample space (S) = 12+x
∴n (S) = 12+x
Let, the event of getting a red ball be A.
∴n(A) = x (Given)
∴ P (A) = n(A)/n(S) = x/x+12
P (a red ball) = x/x+12
(ii) 8 more red balls are added.
∴n (A) = x + 8
n (S) = x + 12 + 8 = x + 20
∴ New probability P(A) = x + 8/x + 20
Now, x+8/x+20 = 2 (x/x+12)
2x (x + 20) = (x + 8) (x + 12)
2 x^{2} + 40x = x^{2} + 8x + 12x + 96
2x^{2} – x^{2} – 20x + 40x – 96 = 0
x^{2} + 20x – 96 = 0
x^{2} + 24x – 4x – 96 = 0
x (x + 24) – 4 (x + 24) = 0
(x + 24) (x – 4) = 0
x + 24 = 0
x = 24
x≠ 24
x – 4 = 0
x = 4
[Number & ball not negative]
∴ Probability of getting red balls
P (A) = x/x+12 = 4/4+12 = 4/16 = 1/4
Thus, P (A) = 1/4 and x = 4.
(7) Two unbiased dice are rolled once. Find the probability of getting
(i) A doublet (equal numbers on both dice)
(ii) The product as a prime number
(iii) The sum as a prime number
(iv) The sum as 1
Solution:
Sample Space {(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
(i) Let, the event of getting doublet be A,
A_{1} = {1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n (A_{1}) = 6
P (A_{1}) = n(A)/n(S) = 6/36 = 1/6
∴ P (doublet) = 1/6
(ii) Let, A_{2} be the event of getting a product as a prime number.
∴ A_{2} = {(1, 2), (1, 3), (1, 5), (2, 1), (3, 1), (5, 1)}
n (A_{2}) = 6
∴ P (A_{2}) = n(A_{2})/n(s) = 6/36 = 1/6
P (the product as a prime number) = 1/6
(iii) Let A_{3} be the event of getting the sum as prime number
A_{3} = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
n (A_{3}} = 15
P (A_{3}) = n(A_{3})/n(s) = 15/36 = 5/12
P (The sum as a prime number) = 5/12
(iv) Let, A_{4} be the event of getting the sum as 1.
A_{4} = {0}
n (A_{4}) = 0
P (A_{4}) = n(A_{4})/n(s) = 0/36 = 0
P (The sum as 1) = 0
(8) Three fair coins are tossed together. Find the probability of getting
(i) All heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Sample Space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴n (s) = 8
(i) Let A_{1} be the event of getting all heads.
∴ A_{1} = {HHH}
n (A_{1}) = 1
∴ P (A_{1}) = n (A_{1})/n(S) = 1/8
P (All heads) = 1/8
(ii) Let A_{2} be the event of getting atleast one tail.
A_{2} = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n (A_{2}) = 7
∴ P (A_{2}) = n(A_{2})/n(S) = 7/8
∴ P (atleast one tail)= 7/8
(iii) Let, A_{3} be the event of getting almost one head.
∴ A_{3} = {HTT, THT, TTH, TTT}
n (A_{3}) = 4
∴ P (A_{3}) = n(A_{3})/n(S) = 4/8 = 1/2
P (Atleast one head) = 1/2
(iv) Let A_{4} be the event of getting almost two tails.
A_{4} = {HTT, TTT, TTH, THT, THH, HHT, HTH}
n (A_{4}) = 7
P (A_{4}) = n(A_{4})/n(S) = 7/8
∴ P (atmost two tails) = 7/8
(9) A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) White
(ii) Black or red
(iii) Not white
(iv) Neither white nor black
Solution:
Given that, Red balls = 5
White balls = 6
Green balls = 7
Black balls = 8
∴n (S) = 5+6+7+8 = 26
(i) Let, A_{1} be the event of getting a white ball.
∴n (A_{1}) = 6
P (A_{1}) = n(A_{1})/n(S) = 6/26 = 3/13
∴ P (White) = 3/13
(ii) Let, A_{2} be the event of getting a black ball.
∴n (A_{2}) = 8
∴ P (A_{2}) = n(A_{2})/n(S) = 8/26
∴ P (Black) = 8/26
Let A_{3} be the event of getting a red ball.
∴n (A_{3}) = 5
P (A_{3}) = n(A_{2})/n(S) = 5/26
Now probability of one black or red balls.
P (A_{2}UA_{3}) = P (A_{2}) + P (A_{3})
= 8/28 + 5/26
= 13/26
= 1/2
Thus, the probability of on black or red ball is ½.
(iii) Let A_{4} be the event of white ball.
∴n (A_{4}) = 6
∴ P (A_{4}) = n(A_{4})/n(S) = 6/26 = 3/13
∴ A_{4} be the event of getting not white.
∴We know that P (A_{4}) + P (A_{4}) = 1
3/13 + P (A_{4}) = 1
∴ P (A_{4}) = 1 – 3/13 = 10/13
∴ P (Not white) = 10/13
(iv) Now, P (A_{4}) = 3/13 [A_{4} = the event of getting white]
P (A_{2}) = 8/26 [A_{2} = the event of getting black]
∴ Probability of getting white or black ball
= P (A_{4}∪ A_{2}) = P (A_{4}) + P (A_{2})
= 3/13 + 8/26
= 6+8/26 = 14/26
∴ Probability of getting neither white nor black ball = 1 – 14/26
= 2614/26
= 12/26
= 6/13
Thus, the probability of neither white nor black is 6/13
(10) In a box there are 20 nondefective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs.
Solution:
Let the number of defective bulbs is x
Sample space = x + 20 [∵ 20 = nondefective]
∴ n (S) = x + 20
Let, A be the event of getting a bulb defective
∴ n (A) = x
P (A) = n(A)/n(S) = x/x+20
Now, given that P(A) = 3/8
x/x+20 = 3/8
8x = 3x + 60
8x – 3x = 60
5x = 60
x = 12
Thus, the number of defective bulbs is 12.
(11) The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Solution:
King diamond, queen diamond, queen heart, jack heart, jack spade and king spade cards remove.
∴n (S) = 526
= 46
(i) Let, A, be the event of getting a clavor.
∴n (A_{1}) = 13
P (A_{1}) = n(A_{1})/n(S) = 13/46
P (a clavor) = 13/46
(ii) Now, we know that total 52 cards two cards are queen of red card.
Now, this two card remove.
∴ Let, A_{2} be the event of getting a queen of red card.
∴n (A_{2}) = 0
P (A_{2}) = n(A_{2})/n(S) = 0/46 = 0
P (a queen of red card) = 0
(iii) We know that total 52 cards two black king card. Now this two card one king card is remove.
Let, A_{3} be the event of getting a king of black card
∴n (A_{3}) = 21 = 1
∴ P (A_{3}) = n(A_{3})/n(S) = 1/46
∴ P (a king of black card) = 1/46
(12) Some boys are playing a game, in which the stone thrown by them landing in a circular region (given in the figure) is considered as win and landing other than the circular region is considered as loss. What is the probability to win the game? (π = 3.4)
Solution:
Area of rectangle region = 4×3 sq. feet
= 12 sq. feet
∴ Sample space = 12
∴n(S) = 12
∴ Circular radius (r) = 1 feet
Area of circular region = πr^{2}
= π × (1)^{2} sq. feet
= π sq. feet
Let, A be the event of win the game.
∴n (A) = π
P (A) = n(A)/n(S) = π/12
P (A) = 3.14/12 [∵ π = 3.14 (Given)]
P (π) = 314/1200
P (A) = 157/600
Therefore, probability to win the game is 157/600.
(13) Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) The same day
(ii) Different days
(iii) Consecutive days?
Solution:
Sample space = 6×6 [∵ Monday to Saturday are total 6 days]
∴n (S) = 36
(i) Priya and amutthan are visit shop in 6 days.
Let, A_{1} be the event of the same day both visit shop.
∴n (A_{1}) = 6
P (A_{1})= n(A_{1})/n(S) = 6/36 = 1/6
∴ P (Same day) = 1/6
(iii) Priya and Amuthan both visit shope in different day = 36 – 6 = 30.
∴ Let, A_{2} be the event of both the visit shope in different days.
∴n (A_{2}) = 30
P (A_{2}) = n(A_{2})/n(S) = 30/36 = 5/6
P (Different days) = 5/6
(iii) Let, A_{3} be the event of the both visit the shope in consecutive days
∴n (A_{3}) = 5
P (A_{3}) = n(A_{3})/n(S) = 5/36 = 5/36
Thus, P (consecutive days) = 5/36
(14) In a game, the entry fee is ₹150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry . If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution:
Sample space = {HHH, HHT, HTH, HTT, THH, THT TTH, TTT}
n (s) = 8
(i) Let A_{1} be the event of gets double entry fee.
∴ A_{1} = {HHH}
n (A_{1}) = 1
P (A_{1}) = n(A_{1})/n(S) = 1/8
P (Gets double entry fee) = 1/8
(ii) Let A_{2} be the events of just gets her entry fee.
A_{2} = {HHT, HTH, HTT, THH, THT, TTH}
n (A_{2}) = 6
P (A_{2}) = n(A_{2})/n(S) = 6/8 = 3/4
P (Just gets her entry fee) = 3/4
(iii) Let, A_{3} be the event of loses the energy fee.
A_{3} = {TTT}
n (A_{3}) = 1
P (A_{3}) = n(A_{3})/n(S) = 1/8
P (Loses the entry fee) = 1/8
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Exercise 8.3
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Exercise 6.1 Solution 
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