10th Maths Book Back Statistics and Probability Ex 8.3

Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 8 – Statistics and Probability Ex 8.3 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 8 Statistics and Probability consist of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 8.3 Answers in English.




10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 8 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

Chapter 8

Exercise 8.3 Statistics and Probability

1. Write the sample space for tossing three coins using a tree diagram.
Solution:

Samacheer Kalvi 10th Maths Book Back Answers

2. Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using a tree diagram).
Solution:

Samacheer Kalvi 10th Maths Book Back Answers

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(\(\overline{\mathbf{A}}\)) =17 : 15 and n(S) = 640 then find
(i) P(\(\overline{\mathbf{A}}\))
(ii) n(A).
Solution:
P(A): P(\(\overline{\mathbf{A}}\)) = 17 : 15

Samacheer Kalvi 10th Maths Book Back Answers

4. A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
Outcomes {O}: {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
Two consecutive tails {F} : {(HTT), (TTH), (TTT)}
n{F} = 3
n{O} = 8

Samacheer Kalvi 10th Maths Book Back Answers

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?
Answer:
Sample space = {1, 2, 3,… ,1000}
n(S) = 1000
(i) Let A be the event of setting square number greater than 500
A = {529, 576, 625, 676, 729, 784, 841, 900, 961}
n(A) = 9
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{1000}\)
The probability that the first player wins prize = \(\frac{9}{1000}\)
(ii) If the first player wins, the number is excluded for the second player.
n(A) = 8 and n(S) = 999
P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{999}\)
Probability the second player wins a prize = \(\frac{8}{999}\)

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:
12 ➝ blue balls
x ➝ red balls
(i) P (red ball) = \(\frac{x}{x+12}\)
(ii) 8 red balls are added to the bag.
∴ 12 ➝ blue balls
x + 8 ➝ red balls
Samacheer Kalvi 10th Maths Book Back Answers
Given that P(ii) = 2 × P(i)
Samacheer Kalvi 10th Maths Book Back Answers
⇒ (x + 8)(x + 12) = 2x(x + 20)
⇒ (x2 + 20x + 96) = 2x2 + 40x
⇒ x2 + 20x – 96 = 0
⇒ x2 + 24x – 4x – 96 = 0
⇒ x(x + 24) – 4(x + 24) = 0
⇒ (x – 4)(x + 24) = 0
∴ x = 4 (or) x = -24
x cannot be negative ⇒ x = 4
Substituting x = 4 in (i),

Samacheer Kalvi 10th Maths Book Back Answers

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:
Doublet = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6,6)}
Total number of outcomes = 6 × 6
n(S) = 36
Number of favourable outcomes = 6

Samacheer Kalvi 10th Maths Books Back Answers

(ii) Number of favourable outcomes = 6
as favourable outcomes = (1, 2), (2, 1), (1, 3), (3, 1),(1, 5),and (5, 1)

Samacheer Kalvi 10th Maths Books Back Answers

(iii) Sum as prime numbers = {(1, 1), (1, 2), (2, 3), (1, 4), (1, 6), (4, 3), (5, 6)}
Number of favourable outcomes = 7
⇒ Probability = \(\frac{7}{36}\)

(iv) With two dice, minimum sum possible = 2
∴ Prob (sum as 1) = 0 [Impossible event]




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
Answer:
Three fair coins are tossed together
Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting atleast one tail.
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
(iii) Let C be the event of getting atmost one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
\(P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let D be the event of getting atmost two tails.
D = {HTT, TTT, TTH, THT, THH, HHT, HTH}
n(D) = 7
\(P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}\)

9.Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Dice 1
S = {1,2, 3, 4, 5, 6}
Dice 2
S = {1,1,2, 2, 3, 3}
Total possible outcomes when they are rolled

Samacheer Kalvi 10th Maths Books Back Answers

n(S) = 36
Event of sum (2) = A = {(1,1), (1,1)},
n(A) = 2,P(A) = \(\frac{2}{36}\)
Event of sum 3 is B = {(1, 2), (1, 2), (2, 1), (2, 1)}

Samacheer Kalvi 10th Maths Books Back Answers

Event of sum 4 is C= {(1, 3), (1, 3), (2, 2), (2, 2), (3, 1) (3, 1)}
n(C) = 6

Samacheer Kalvi 10th Maths Books Back Answers

Event of getting the sum 5 is
D = {(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)}
n(D) = 6, P(D) = \(\frac{6}{36}\) .
Event of getting the sum 6 is
E = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E) = 6, P(E) = \(\frac{6}{36}\)
Event of getting the sum 7 is
F = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(F) = 6
P(F) = \(\frac{6}{36}\)
Event of getting the sum 8 is
G = {(5, 3), (5, 3), (6, 2), (6, 2)}

Samacheer Kalvi 10th Maths Books Back Answers

Event of getting the sum 9 is
H = {(6, 3), (6, 3), n(H) = 2

Samacheer Kalvi 10th Maths Books Back Answers

10. A bag contains 5 red balls, 6 white balls, 7 green balls, and 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Solution:
5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26

Samacheer Kalvi 10th Maths Books Back Answers

11. In a box there are 20 non-defective 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 be ‘x’
Total number of bulbs = x + 20

Samacheer Kalvi 10th Maths Books Back Answers

⇒ 8x = 3x + 60
⇒ 5x = 60
⇒ x = 12
∴ No.of defective bulbs are = 12.

12. 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:
Samacheer Kalvi 10th Maths Books Back Answers
(i.e) remaining number of cards = 52 – 6 = 46 13
(i) P(a clavor) = \(\frac{13}{46}\)
(ii) P(queen of red card) = 0 as both Queen of diamond and heart have been removed.
(iii) only K of clavor is in the deck
⇒ P(king of black card) = \(\frac{1}{46}\)

13. Some boys are playing a game, in which the stone was thrown by them landing in a circular region (given in the figure) is considered as a win, and landing other than the circular region is considered a loss. What is the probability to win the game?

Solution:

Samacheer Kalvi 10th Maths Books Back Answers

14. 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:

Samacheer Kalvi 10th Maths Books Back Answers

15. 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 a double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution:

Samacheer Kalvi 10th Maths Books Back Answers

Other Important Links for 10th Maths Book Back Answers/Solutions:

Click Here for Complete Samacheer Kalvi 10th Book Back Solution Guide PDF – Samacheer Kalvi 10th std Maths Book Back Answers