Samacheer Kalvi 10th Maths Exercise 8.3 Solutions

10th Maths Exercise 8.3 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 29 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance28 answers · full steps below

Q1 $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ ; $n(S)=8$
Q3 (i) $P(\overline{A})=\dfrac{15}{32}$
Q3 (ii) $n(A)=340$
Q4 $\dfrac{3}{8}$
Q5 (i) $\dfrac{9}{1000}$
Q5 (ii) $\dfrac{8}{999}$
Q6 (i) $P(\text{red})=\dfrac{x}{12+x}$ ; with $x=4$ (from part (ii)) this equals $\dfrac{1}{4}$ .
Q6 (ii) $x=4$
Q7 (i) $\dfrac{1}{6}$
Q7 (ii) $\dfrac{1}{6}$
Q7 (iii) $\dfrac{5}{12}$
Q7 (iv) $0$
Q8 (i) $\dfrac{1}{8}$
Q8 (ii) $\dfrac{7}{8}$
Q8 (iii) $\dfrac{1}{2}$
Q8 (iv) $\dfrac{7}{8}$
Q9 (i) $\dfrac{3}{13}$
Q9 (ii) $\dfrac{1}{2}$
Q9 (iii) $\dfrac{10}{13}$
Q9 (iv) $\dfrac{6}{13}$
Q10Number of defective bulbs $=12$
Q11 $\dfrac{157}{600}\approx 0.2617$
Q12 (i) $\dfrac{1}{6}$
Q12 (ii) $\dfrac{5}{6}$
Q12 (iii) $\dfrac{5}{18}$
Q13 (i) $\dfrac{1}{8}$
Q13 (ii) $\dfrac{3}{4}$
Q13 (iii) $\dfrac{1}{8}$

Q1. Write the sample space for tossing three coins using tree diagram.

Each coin can show head $H$ or tail $T$ .
Tree: first coin gives 2 branches $H,T$ ; each splits into $H,T$ for the second coin; each of those splits again for the third coin.
So $2\times2\times2=8$ end branches.
Reading along the branches,
$S=\{HHH,\,HHT,\,HTH,\,HTT,\,THH,\,THT,\,TTH,\,TTT\}$

Answer

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ ; $n(S)=8$

Q2. Write the sample space for selecting two balls at a time from …

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

Show steps (9)

First ball can be any of $1,2,3,4,5,6$ — 6 branches.
Since two balls are taken at a time, the second ball must be different from the first — each branch splits into 5.
So $n(S)=6\times5=30$ .
$S=\{(1,2),(1,3),(1,4),(1,5),(1,6),$
$(2,1),(2,3),(2,4),(2,5),(2,6),$
$(3,1),(3,2),(3,4),(3,5),(3,6),$
$(4,1),(4,2),(4,3),(4,5),(4,6),$
$(5,1),(5,2),(5,3),(5,4),(5,6),$
$(6,1),(6,2),(6,3),(6,4),(6,5)\}$

Answer

$S=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$ ; $n(S)=30$

Q3. If $A$ is an event of a random experiment such that $P(A) : P(\overline{A}) = 17{:}15$ …

If $A$ is an event of a random experiment such that $P(A) : P(\overline{A}) = 17{:}15$ and $n(S)=640$ then find

(i)

$P(\overline{A})$ .

Let $P(A)=17k$ and $P(\overline{A})=15k$ .
$P(A)+P(\overline{A})=1$
$17k+15k=1\Rightarrow 32k=1\Rightarrow k=\dfrac{1}{32}$
$P(\overline{A})=15\times\dfrac{1}{32}=\dfrac{15}{32}$

Answer

$P(\overline{A})=\dfrac{15}{32}$

(ii)

$n(A)$ .

From (i), $P(A)=\dfrac{17}{32}$ .
$P(A)=\dfrac{n(A)}{n(S)}$
$\dfrac{17}{32}=\dfrac{n(A)}{640}$
$n(A)=\dfrac{17}{32}\times640=340$

Answer

$n(A)=340$

Q4. A coin is tossed thrice. What is the probability of getting two …

A coin is tossed thrice. What is the probability of getting two consecutive tails?

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ ; $n(S)=8$
Let $A$ be the event of getting two consecutive tails.
$A=\{HTT,TTH,TTT\}$ ; $n(A)=3$
$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{8}$

Answer

$\dfrac{3}{8}$

Q5. At a fete, cards bearing numbers 1 to 1000, one number on …

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?

$n(S)=1000$
Perfect squares greater than $500$ and up to $1000$ :
$23^2=529,\;24^2=576,\;25^2=625,\;26^2=676,\;27^2=729,\;28^2=784,\;29^2=841,\;30^2=900,\;31^2=961$
Let $A$ be the event that the first player wins; $n(A)=9$
$P(A)=\dfrac{9}{1000}$

Answer

$\dfrac{9}{1000}$

(ii)

the second player wins a prize, if the first has won?

The first player won, so one winning card is removed and not replaced.
Remaining cards $=1000-1=999$ ; remaining winning numbers $=9-1=8$
$P(\text{second player wins})=\dfrac{8}{999}$

Answer

$\dfrac{8}{999}$

Q6. A bag contains $12$ blue balls and $x$ red balls. If one …

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?

Show steps (6)

Total number of balls $n(S)=12+x$ (12 blue and $x$ red).
Let $R$ be the event of drawing a red ball, so $n(R)=x$ .
$P(R)=\dfrac{n(R)}{n(S)}=\dfrac{x}{12+x}$ .
This is the required answer in terms of $x$ ; the value of $x$ is found only in part (ii).
From part (ii): $\dfrac{x+8}{20+x}=2\left(\dfrac{x}{12+x}\right)\Rightarrow x^{2}+20x-96=0\Rightarrow (x+24)(x-4)=0$ , so $x=4$ (rejecting $x=-24$ ).
Hence, substituting $x=4$ , the numerical value of the probability in (i) is $\dfrac{4}{12+4}=\dfrac{4}{16}=\dfrac{1}{4}$ , which matches the printed answer.

Answer

$P(\text{red})=\dfrac{x}{12+x}$ ; with $x=4$ (from part (ii)) this equals $\dfrac{1}{4}$ .

(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 $x$ .

Show steps (7)

After adding 8 red balls: red $=x+8$ , total $=12+x+8=20+x$
Given, $\dfrac{x+8}{20+x}=2\left(\dfrac{x}{12+x}\right)$
$(x+8)(12+x)=2x(20+x)$
$x^2+20x+96=2x^2+40x$
$x^2+20x-96=0$
$(x+24)(x-4)=0$
$x=4$ (since $x$ cannot be negative)

Answer

$x=4$

Q7. Two unbiased dice are rolled once. Find the probability of getting

(i)

a doublet (equal numbers on both dice).

Two dice: $n(S)=36$
Let $A$ be the event of getting a doublet.
$A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$ ; $n(A)=6$
$P(A)=\dfrac{6}{36}=\dfrac{1}{6}$

Answer

$\dfrac{1}{6}$

(ii)

the product as a prime number.

Product is prime only when one die shows $1$ and the other shows a prime.
$B=\{(1,2),(1,3),(1,5),(2,1),(3,1),(5,1)\}$ ; $n(B)=6$
$P(B)=\dfrac{6}{36}=\dfrac{1}{6}$

Answer

$\dfrac{1}{6}$

(iii)

the sum as a prime number.

Possible prime sums are $2,3,5,7,11$ .
$C=\{(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(5,6),(6,5)\}$
$n(C)=15$
$P(C)=\dfrac{15}{36}=\dfrac{5}{12}$

Answer

$\dfrac{5}{12}$

(iv)

the sum as 1.

The least possible sum with two dice is $1+1=2$ .
So no outcome gives sum $1$ ; it is an impossible event.
$n(D)=0\Rightarrow P(D)=\dfrac{0}{36}=0$

Answer

$0$

Q8. Three fair coins are tossed together. Find the probability of getting

(i)

all heads.

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ ; $n(S)=8$
$A=\{HHH\}$ ; $n(A)=1$
$P(A)=\dfrac{1}{8}$

Answer

$\dfrac{1}{8}$

(ii)

atleast one tail.

'Atleast one tail' is the complement of 'no tail' i.e. $HHH$ .
$P(\text{no tail})=\dfrac{1}{8}$
$P(\text{atleast one tail})=1-\dfrac{1}{8}=\dfrac{7}{8}$

Answer

$\dfrac{7}{8}$

(iii)

atmost one head.

Atmost one head means $0$ head or $1$ head.
$C=\{TTT,HTT,THT,TTH\}$ ; $n(C)=4$
$P(C)=\dfrac{4}{8}=\dfrac{1}{2}$

Answer

$\dfrac{1}{2}$

(iv)

atmost two tails.

Atmost two tails is the complement of 'three tails' i.e. $TTT$ .
$P(TTT)=\dfrac{1}{8}$
$P(\text{atmost two tails})=1-\dfrac{1}{8}=\dfrac{7}{8}$

Answer

$\dfrac{7}{8}$

Q9. A bag contains 5 red balls, 6 white balls, 7 green balls …

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.

Total balls $n(S)=5+6+7+8=26$
$n(\text{white})=6$
$P(\text{white})=\dfrac{6}{26}=\dfrac{3}{13}$

Answer

$\dfrac{3}{13}$

(ii)

black or red.

$n(\text{black or red})=8+5=13$
$P=\dfrac{13}{26}=\dfrac{1}{2}$

Answer

$\dfrac{1}{2}$

(iii)

not white.

$P(\text{white})=\dfrac{3}{13}$
$P(\text{not white})=1-\dfrac{3}{13}=\dfrac{10}{13}$

Answer

$\dfrac{10}{13}$

(iv)

neither white nor black.

Neither white nor black means red or green.
$n=5+7=12$
$P=\dfrac{12}{26}=\dfrac{6}{13}$

Answer

$\dfrac{6}{13}$

Q10. In a box there are $20$ non-defective and some defective bulbs. If …

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 $\dfrac{3}{8}$ then, find the number of defective bulbs.

Show steps (6)

Let the number of defective bulbs be $x$ .
Total bulbs $=20+x$
$P(\text{defective})=\dfrac{x}{20+x}=\dfrac{3}{8}$
$8x=3(20+x)$
$8x=60+3x\Rightarrow 5x=60$
$x=12$

Answer

Number of defective bulbs $=12$

Q11. Some boys are playing a game, in which the stone thrown by …

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? ( $\pi = 3.14$ )

A shaded rectangle 4 feet wide and 3 feet tall with a circle inside whose radius is marked 1 feet.

Total region = rectangle of area $=4\times3=12$ sq.feet
Winning region = circle of radius $1$ ft, area $=\pi r^2=3.14\times1^2=3.14$ sq.feet
$P(\text{win})=\dfrac{\text{area of circle}}{\text{area of rectangle}}=\dfrac{3.14}{12}$
$=\dfrac{314}{1200}=\dfrac{157}{600}$

Answer

$\dfrac{157}{600}\approx 0.2617$

Q12. Two customers Priya and Amuthan are visiting a particular shop in the …

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?

Each can visit on any of 6 days (Mon–Sat).
$n(S)=6\times6=36$
Same day: $\{(M,M),(T,T),(W,W),(Th,Th),(F,F),(S,S)\}$ ; $n(A)=6$
$P(A)=\dfrac{6}{36}=\dfrac{1}{6}$

Answer

$\dfrac{1}{6}$

(ii)

different days?

Different days is the complement of same day.
$P=1-\dfrac{1}{6}=\dfrac{5}{6}$

Answer

$\dfrac{5}{6}$

(iii)

consecutive days?

Consecutive day pairs: (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat) — 5 pairs.
Each pair can occur in 2 orders, so $n(C)=5\times2=10$
$P(C)=\dfrac{10}{36}=\dfrac{5}{18}$

Answer

$\dfrac{5}{18}$

Q13. In a game, the entry fee is ₹ $150$ . The game consists of …

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.

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ ; $n(S)=8$
She gets double the entry fee only for 3 heads.
$A=\{HHH\}$ ; $n(A)=1$
$P(A)=\dfrac{1}{8}$

Answer

$\dfrac{1}{8}$

(ii)

just gets her entry fee.

She just gets back the entry fee when one or two heads show.
$B=\{HTT,THT,TTH,HHT,HTH,THH\}$ ; $n(B)=6$
$P(B)=\dfrac{6}{8}=\dfrac{3}{4}$

Answer

$\dfrac{3}{4}$

(iii)

loses the entry fee.

She loses when no head shows.
$C=\{TTT\}$ ; $n(C)=1$
$P(C)=\dfrac{1}{8}$

Answer

$\dfrac{1}{8}$

Quick check

The probability of getting a job for a person is $\dfrac{x}{3}$ . If the probability of not getting the job is $\dfrac{2}{3}$ then the value of $x$ is
1. (1) $2$
2. (2) $1$
3. (3) $3$
4. (4) $1.5$
Show answer
Answer: (2) $1$
1. $P(A)+P(\bar{A})=1$
2. $\dfrac{x}{3}+\dfrac{2}{3}=1$
3. $\dfrac{x}{3}=\dfrac{1}{3}\Rightarrow x=1$
Question 1 of 5
The mean of $100$ observations is $40$ and their standard deviation is $3$ . The sum of squares of all observations is
1. (1) $40000$
2. (2) $160900$
3. (3) $160000$
4. (4) $30000$
Show answer
Answer: (2) $160900$
1. $\sigma^2=\dfrac{\sum x^2}{n}-\left(\bar{x}\right)^2$
2. $9=\dfrac{\sum x^2}{100}-1600$
3. $\dfrac{\sum x^2}{100}=1609$
4. $\sum x^2=160900$
Question 2 of 5
Which of the following is incorrect?
1. (1) $P(A) > 1$
2. (2) $0 \leq P(A) \leq 1$
3. (3) $P(\phi) = 0$
4. (4) $P(A) + P(\bar{A}) = 1$
Show answer
Answer: (1) $P(A) > 1$
1. Probability of any event always lies between $0$ and $1$ .
2. So $P(A)>1$ is impossible.
Question 3 of 5
The probability a red marble selected at random from a jar containing $p$ red, $q$ blue and r green marbles is
1. (1) $\dfrac{q}{p+q+r}$
2. (2) $\dfrac{p}{p+q+r}$
3. (3) $\dfrac{p+q}{p+q+r}$
4. (4) $\dfrac{p+r}{p+q+r}$
Show answer
Answer: (2) $\dfrac{p}{p+q+r}$
1. Total marbles $=p+q+r$ ; red marbles $=p$ .
2. $P(\text{red})=\dfrac{p}{p+q+r}$
Question 4 of 5
The range of the data $8, 8, 8, 8, 8. . . 8$ is
1. (1) $0$
2. (2) $1$
3. (3) $8$
4. (4) $3$
Show answer
Answer: (1) $0$
1. Largest value $= 8$ , smallest value $= 8$ .
2. Range $= L - S = 8 - 8 = 0$ .
Question 5 of 5

Chapter 8 formulas

Range
$R = L - S \quad (L\text{ - largest value},\ S\text{ - smallest value})$
Coefficient of range
$\text{Coefficient of range} = \frac{L-S}{L+S}$
Variance
$\sigma^{2} = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^{2}}{n}$
Standard deviation
$\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^{2}}{n}}$
Standard deviation (ungrouped data) - Direct method
$\sigma = \sqrt{\frac{\sum x_i^{2}}{n} - \left(\frac{\sum x_i}{n}\right)^{2}}$
Standard deviation (ungrouped data) - Mean method
$\sigma = \sqrt{\frac{\sum d_i^{2}}{n}}$
Standard deviation (ungrouped data) - Assumed mean method
$\sigma = \sqrt{\frac{\sum d_i^{2}}{n} - \left(\frac{\sum d_i}{n}\right)^{2}}$
Standard deviation (ungrouped data) - Step deviation method
$\sigma = c \times \sqrt{\frac{\sum d_i^{2}}{n} - \left(\frac{\sum d_i}{n}\right)^{2}}$
Standard deviation of first n natural numbers
$\sigma = \sqrt{\frac{n^{2}-1}{12}}$
Standard deviation (grouped data) - Mean method
$\sigma = \sqrt{\frac{\sum f_i d_i^{2}}{N}}$
Standard deviation (grouped data) - Assumed mean method
$\sigma = \sqrt{\frac{\sum f_i d_i^{2}}{N} - \left(\frac{\sum f_i d_i}{N}\right)^{2}}$
Standard deviation (grouped data) - Step deviation method
$\sigma = C \times \sqrt{\frac{\sum f_i d_i^{2}}{N} - \left(\frac{\sum f_i d_i}{N}\right)^{2}}$
Coefficient of variation
$\text{C.V} = \frac{\sigma}{\bar{x}} \times 100\%$
Consistency using C.V.
$\text{Smaller C.V.} \Rightarrow \text{more consistent data};\quad \text{larger C.V.} \Rightarrow \text{less consistent data}$
Mutually exclusive events
$A,\ B \text{ are mutually exclusive if } A \cap B = \phi$
Probability of an event
$P(E) = \frac{n(E)}{n(S)}$
Sure and impossible events
$P(S) = 1,\qquad P(\phi) = 0$
Range of probability
$0 \le P(E) \le 1$
Complement of an event
$P(\bar{E}) = 1 - P(E)$
Addition theorem (mutually exclusive events)
$P(A \cup B) = P(A) + P(B)$
Probability of only A
$P(A \cap \bar{B}) = P(\text{only } A) = P(A) - P(A \cap B)$
Probability of only B
$P(\bar{A} \cap B) = P(\text{only } B) = P(B) - P(A \cap B)$
Addition theorem for any two events
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Addition theorem for any three events
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

Spotted a mistake? Email us at [email protected] and we will fix it.

Q1. Write the sample space for tossing three coins using tree diagram.

Q2. Write the sample space for selecting two balls at a time from …

Q3. If AAA is an event of a random experiment such that P(A):P(A‾)=17:15P(A) : P(\overline{A}) = 17{:}15P(A):P(A)=17:15 …

(i)

(ii)

Q4. A coin is tossed thrice. What is the probability of getting two …

Q5. At a fete, cards bearing numbers 1 to 1000, one number on …

(i)

(ii)

Q6. A bag contains 121212 blue balls and xxx red balls. If one …

(i)

(ii)

Q7. Two unbiased dice are rolled once. Find the probability of getting

(i)

(ii)

(iii)

(iv)

Q8. Three fair coins are tossed together. Find the probability of getting

(i)

(ii)

(iii)

(iv)

Q9. A bag contains 5 red balls, 6 white balls, 7 green balls …

(i)

(ii)

(iii)

(iv)

Q10. In a box there are 202020 non-defective and some defective bulbs. If …

Q11. Some boys are playing a game, in which the stone thrown by …

Q12. Two customers Priya and Amuthan are visiting a particular shop in the …

(i)

(ii)

(iii)

Q13. In a game, the entry fee is ₹150150150. The game consists of …

(i)

(ii)

(iii)

Quick check

Q3. If $A$ is an event of a random experiment such that $P(A) : P(\overline{A}) = 17{:}15$ …

Q6. A bag contains $12$ blue balls and $x$ red balls. If one …

Q10. In a box there are $20$ non-defective and some defective bulbs. If …

Q13. In a game, the entry fee is ₹ $150$ . The game consists of …