Samacheer kalvi 10th Maths - Statistics and Probability Ex 8.3

10th Maths Book Back Question and Answers – Chapter 8 Exercise 8.3:

Samacheer Kalvi 10th Standard Maths Book Back Questions with Answers PDF uploaded and the same given below. Class-tenth candidates and those preparing for TNPSC exams can check the Maths Book Back Answers PDF below. Samacheer Kalvi Class 10th Std Maths Book Back Answers Chapter 8 Statistics and Probability Exercise 8.3 Solutions are available below. Check the complete Samacheer Kalvi 10th Maths – Statistics and Probability Ex 8.3 Book Back Answers below:

We also provide class 10th other units Maths Book Back One, Two, and Five Mark Solutions Guide on our site. Students looking for the 10th standard Maths Ex 8.3 – Statistics and Probability Book Back Questions with Answer PDF

For the complete Samacheer Kalvi 10th Maths Book Back Solutions Guide PDF, check the link – Samacheer Kalvi 10th Maths Book Back Answers

Samacheer Kalvi 10th Maths Book Back Answers – Ex 8.3 Statistics and Probability

Samacheer Kalvi 10th Maths Book Subject One Mark, Two Mark, Five Mark Guide questions and answers are below. Check Maths Book Back Questions with Answers. Take the printout and use it for exam purposes.

For Samacheer Kalvi 10th Maths Book PDF, check the link – 10th Maths Book PDF

Chapter 8 Exercise 8.3 Statistics and Probability

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

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

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

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

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

Given that P(ii) = 2 × P(i)

⇒ (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),

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

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

(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

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)}

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

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)}

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

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

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

⇒ 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:

(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:

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:

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:

How To Get Govt. Job Without Wasting Years?
CLICK HERE

07 Apr 2022