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Samacheer Kalvi 10th Maths Exercise 8.4 Solutions

10th Maths Exercise 8.4 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1P(A∩B)=1115P(A\cap B)=\dfrac{11}{15}
  2. Q2 (i)P(A‾)=0.58P(\overline{A})=0.58
  3. Q2 (ii)P(B‾)=0.52P(\overline{B})=0.52
  4. Q2 (iii)P(A or B)=0.74P(A\text{ or }B)=0.74
  5. Q3P(B)=0.1P(B)=0.1
  6. Q4P(A‾)+P(B‾)=1.2P(\overline{A})+P(\overline{B})=1.2
  7. Q50.20.2
  8. Q659\dfrac{5}{9}
  9. Q71318\dfrac{13}{18}
  10. Q878\dfrac{7}{8}
  11. Q9P(A∩B)=73280P(A\cap B)=\dfrac{73}{280}
  12. Q101740\dfrac{17}{40}
  13. Q1111
  14. Q12P(A)=1148, P(B)=1124, P(C)=1116P(A)=\dfrac{11}{48},\ P(B)=\dfrac{11}{24},\ P(C)=\dfrac{11}{16}
  15. Q132935\dfrac{29}{35}

Q1. If P(A)=23P(A) = \dfrac{2}{3} , P(B)=25P(B) = \dfrac{2}{5} , P(A∪B)=13P(A \cup B) = \dfrac{1}{3} then find P(A∩B)P(A \cap B) .

  1. P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)

  2. P(A∩B)=P(A)+P(B)−P(A∪B)P(A\cap B)=P(A)+P(B)-P(A\cup B)

  3. =23+25−13=\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{1}{3}

  4. =10+6−515=1115=\dfrac{10+6-5}{15}=\dfrac{11}{15}

Answer

P(A∩B)=1115P(A\cap B)=\dfrac{11}{15}

Q2. AA and BB are two events such that, P(A)=0.42P(A) = 0.42 , P(B)=0.48P(B) = 0.48 …

AA and BB are two events such that, P(A)=0.42P(A) = 0.42, P(B)=0.48P(B) = 0.48, and P(A∩B)=0.16P(A \cap B) = 0.16. Find

(i)

P(not A)P(\text{not } A)

  1. P(not A)=1−P(A)P(\text{not }A)=1-P(A)

  2. =1−0.42=0.58=1-0.42=0.58

Answer

P(A‾)=0.58P(\overline{A})=0.58

(ii)

P(not B)P(\text{not } B)

  1. P(not B)=1−P(B)P(\text{not }B)=1-P(B)

  2. =1−0.48=0.52=1-0.48=0.52

Answer

P(B‾)=0.52P(\overline{B})=0.52

(iii)

P(A or B)P(A \text{ or } B)

  1. P(A or B)=P(A∪B)=P(A)+P(B)−P(A∩B)P(A\text{ or }B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)

  2. =0.42+0.48−0.16=0.42+0.48-0.16

  3. =0.74=0.74

Answer

P(A or B)=0.74P(A\text{ or }B)=0.74

Q3. If AA and BB are two mutually exclusive events of a random …

If AA and BB are two mutually exclusive events of a random experiment and P(not A)=0.45P(\text{not } A) = 0.45, P(A∪B)=0.65P(A \cup B) = 0.65, then find P(B)P(B).

  1. P(A)=1−P(A‾)=1−0.45=0.55P(A)=1-P(\overline{A})=1-0.45=0.55

  2. AA and BB are mutually exclusive, so P(A∩B)=0P(A\cap B)=0

  3. P(A∪B)=P(A)+P(B)P(A\cup B)=P(A)+P(B)

  4. 0.65=0.55+P(B)0.65=0.55+P(B)

  5. P(B)=0.10P(B)=0.10

Answer

P(B)=0.1P(B)=0.1

Q4. The probability that atleast one of AA and BB occur is 0.60.6 …

The probability that atleast one of AA and BB occur is 0.60.6. If AA and BB occur simultaneously with probability 0.20.2, then find P(A‾)+P(B‾)P(\overline{A}) + P(\overline{B}).

  1. P(A∪B)=0.6P(A\cup B)=0.6, P(A∩B)=0.2P(A\cap B)=0.2

  2. P(A)+P(B)=P(A∪B)+P(A∩B)=0.6+0.2=0.8P(A)+P(B)=P(A\cup B)+P(A\cap B)=0.6+0.2=0.8

  3. P(A‾)+P(B‾)=(1−P(A))+(1−P(B))P(\overline{A})+P(\overline{B})=\big(1-P(A)\big)+\big(1-P(B)\big)

  4. =2−0.8=1.2=2-0.8=1.2

Answer

P(A‾)+P(B‾)=1.2P(\overline{A})+P(\overline{B})=1.2

Q5. The probability of happening of an event AA is 0.50.5 and that …

The probability of happening of an event AA is 0.50.5 and that of BB is 0.30.3. If AA and BB are mutually exclusive events, then find the probability that neither AA nor BB happen.

  1. A,BA,B mutually exclusive ⇒P(A∪B)=P(A)+P(B)=0.5+0.3=0.8\Rightarrow P(A\cup B)=P(A)+P(B)=0.5+0.3=0.8

  2. P(neither AA nor BB) =P(A‾∩B‾)=P(A∪B‾)=P(\overline{A}\cap\overline{B})=P(\overline{A\cup B})

  3. =1−P(A∪B)=1−0.8=0.2=1-P(A\cup B)=1-0.8=0.2

Answer

0.20.2

Q6. Two dice are rolled once. Find the probability of getting an even …

Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 88.

  1. Two dice: n(S)=36n(S)=36

  2. AA = even number on first die ⇒n(A)=3×6=18\Rightarrow n(A)=3\times 6=18

  3. BB = face sum 88 = {(2,6),(3,5),(4,4),(5,3),(6,2)}\{(2,6),(3,5),(4,4),(5,3),(6,2)\}, n(B)=5n(B)=5

  4. A∩B={(2,6),(4,4),(6,2)}A\cap B=\{(2,6),(4,4),(6,2)\}, n(A∩B)=3n(A\cap B)=3

  5. P(A∪B)=1836+536−336=2036=59P(A\cup B)=\dfrac{18}{36}+\dfrac{5}{36}-\dfrac{3}{36}=\dfrac{20}{36}=\dfrac{5}{9}

Answer

59\dfrac{5}{9}

Q7. A box contains cards numbered 3, 5, 7, 9, … 35, 373,\ 5,\ 7,\ 9,\ \ldots\ 35,\ 37 . A card is drawn at …

A box contains cards numbered 3, 5, 7, 9, … 35, 373,\ 5,\ 7,\ 9,\ \ldots\ 35,\ 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 77 or a prime number.

  1. Cards are odd numbers 3,5,7,…,373,5,7,\dots,37, so n(S)=37−32+1=18n(S)=\dfrac{37-3}{2}+1=18

  2. AA = multiples of 77 = {7,21,35}\{7,21,35\}, n(A)=3n(A)=3

  3. BB = primes = {3,5,7,11,13,17,19,23,29,31,37}\{3,5,7,11,13,17,19,23,29,31,37\}, n(B)=11n(B)=11

  4. A∩B={7}A\cap B=\{7\}, n(A∩B)=1n(A\cap B)=1

  5. P(A∪B)=318+1118−118=1318P(A\cup B)=\dfrac{3}{18}+\dfrac{11}{18}-\dfrac{1}{18}=\dfrac{13}{18}

Answer

1318\dfrac{13}{18}

Q8. Three unbiased coins are tossed once. Find the probability of getting atmost …

Three unbiased coins are tossed once. Find the probability of getting atmost 22 tails or atleast 22 heads.

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

  2. AA = atmost 22 tails = all except TTTTTT, n(A)=7n(A)=7

  3. BB = atleast 22 heads = {HHH,HHT,HTH,THH}\{HHH,HHT,HTH,THH\}, n(B)=4n(B)=4

  4. A∩B={HHH,HHT,HTH,THH}A\cap B=\{HHH,HHT,HTH,THH\}, n(A∩B)=4n(A\cap B)=4

  5. P(A∪B)=78+48−48=78P(A\cup B)=\dfrac{7}{8}+\dfrac{4}{8}-\dfrac{4}{8}=\dfrac{7}{8}

Answer

78\dfrac{7}{8}

Q9. The probability that a person will get an electrification contract is 35\dfrac{3}{5} …

The probability that a person will get an electrification contract is 35\dfrac{3}{5} and the probability that he will not get plumbing contract is 58\dfrac{5}{8}. The probability of getting atleast one contract is 57\dfrac{5}{7}. What is the probability that he will get both?

  1. Let AA = getting electrification contract, BB = getting plumbing contract

  2. P(A)=35P(A)=\dfrac{3}{5}, P(B‾)=58⇒P(B)=1−58=38P(\overline{B})=\dfrac{5}{8}\Rightarrow P(B)=1-\dfrac{5}{8}=\dfrac{3}{8}

  3. Atleast one contract: P(A∪B)=57P(A\cup B)=\dfrac{5}{7}

  4. P(A∩B)=P(A)+P(B)−P(A∪B)P(A\cap B)=P(A)+P(B)-P(A\cup B)

  5. =35+38−57=168+105−200280=73280=\dfrac{3}{5}+\dfrac{3}{8}-\dfrac{5}{7}=\dfrac{168+105-200}{280}=\dfrac{73}{280}

Answer

P(A∩B)=73280P(A\cap B)=\dfrac{73}{280}

Q10. In a town of 80008000 people, 13001300 are over 5050 years and …

In a town of 80008000 people, 13001300 are over 5050 years and 30003000 are females. It is known that 30%30\% of the females are over 5050 years. What is the probability that a chosen individual from the town is either a female or over 5050 years?

Show steps (6)

  1. n(S)=8000n(S)=8000

  2. AA = over 5050 years, n(A)=1300n(A)=1300

  3. BB = female, n(B)=3000n(B)=3000

  4. n(A∩B)=30%n(A\cap B)=30\% of 3000=9003000=900

  5. P(A∪B)=13008000+30008000−9008000P(A\cup B)=\dfrac{1300}{8000}+\dfrac{3000}{8000}-\dfrac{900}{8000}

  6. =34008000=1740=\dfrac{3400}{8000}=\dfrac{17}{40}

Answer

1740\dfrac{17}{40}

Q11. A coin is tossed thrice. Find the probability of getting exactly two …

A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.

Show steps (7)

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

  2. AA = exactly two heads ={HHT,HTH,THH}=\{HHT,HTH,THH\}, n(A)=3n(A)=3

  3. BB = atleast one tail = all except HHHHHH, n(B)=7n(B)=7

  4. CC = two consecutive heads ={HHH,HHT,THH}=\{HHH,HHT,THH\}, n(C)=3n(C)=3

  5. A∩B={HHT,HTH,THH}A\cap B=\{HHT,HTH,THH\}, n=3n=3; B∩C={HHT,THH}B\cap C=\{HHT,THH\}, n=2n=2; A∩C={HHT,THH}A\cap C=\{HHT,THH\}, n=2n=2

  6. A∩B∩C={HHT,THH}A\cap B\cap C=\{HHT,THH\}, n=2n=2

  7. P(A∪B∪C)=3+7+3−3−2−2+28=88=1P(A\cup B\cup C)=\dfrac{3+7+3-3-2-2+2}{8}=\dfrac{8}{8}=1

Answer

11

Q12. If A, B, CA,\ B,\ C are any three events such that probability of BB is …

If A, B, CA,\ B,\ C are any three events such that probability of BB is twice as that of probability of AA and probability of CC is thrice as that of probability of AA and if P(A∩B)=16P(A \cap B) = \dfrac{1}{6}, P(B∩C)=14P(B \cap C) = \dfrac{1}{4}, P(A∩C)=18P(A \cap C) = \dfrac{1}{8}, P(A∪B∪C)=910P(A \cup B \cup C) = \dfrac{9}{10}, P(A∩B∩C)=115P(A \cap B \cap C) = \dfrac{1}{15}, then find P(A),P(B)P(A), P(B) and P(C)P(C)?

Show steps (6)

  1. Let P(A)=xP(A)=x, then P(B)=2xP(B)=2x, P(C)=3xP(C)=3x

  2. P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)

  3. 910=x+2x+3x−16−14−18+115\dfrac{9}{10}=x+2x+3x-\dfrac{1}{6}-\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{15}

  4. 910=6x−1324+115\dfrac{9}{10}=6x-\dfrac{13}{24}+\dfrac{1}{15}

  5. 6x=910+1324−115=108+65−8120=165120=1186x=\dfrac{9}{10}+\dfrac{13}{24}-\dfrac{1}{15}=\dfrac{108+65-8}{120}=\dfrac{165}{120}=\dfrac{11}{8}

  6. x=1148x=\dfrac{11}{48}

Answer

P(A)=1148, P(B)=1124, P(C)=1116P(A)=\dfrac{11}{48},\ P(B)=\dfrac{11}{24},\ P(C)=\dfrac{11}{16}

Q13. In a class of 3535 , students are numbered from 11 to …

In a class of 3535, students are numbered from 11 to 3535. The ratio of boys to girls is 4:34{:}3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.

Show steps (6)

  1. n(S)=35n(S)=35; boys : girls =4:3⇒=4:3\Rightarrow boys =20=20 (roll 11–2020), girls =15=15 (roll 2121–3535)

  2. AA = boy with prime roll ={2,3,5,7,11,13,17,19}=\{2,3,5,7,11,13,17,19\}, n(A)=8n(A)=8

  3. BB = girl with composite roll ={21,22,24,25,26,27,28,30,32,33,34,35}=\{21,22,24,25,26,27,28,30,32,33,34,35\}, n(B)=12n(B)=12

  4. CC = even roll ={2,4,…,34}=\{2,4,\dots,34\}, n(C)=17n(C)=17

  5. A∩B=ϕA\cap B=\phi, n=0n=0; A∩C={2}A\cap C=\{2\}, n=1n=1; B∩C={22,24,26,28,30,32,34}B\cap C=\{22,24,26,28,30,32,34\}, n=7n=7; A∩B∩C=ϕA\cap B\cap C=\phi

  6. P(A∪B∪C)=8+12+17−0−1−7+035=2935P(A\cup B\cup C)=\dfrac{8+12+17-0-1-7+0}{35}=\dfrac{29}{35}

Answer

2935\dfrac{29}{35}

Quick check

  1. The probability of getting a job for a person is x3\dfrac{x}{3}. If the probability of not getting the job is 23\dfrac{2}{3} then the value of xx is

    1. (1) 22
    2. (2) 11
    3. (3) 33
    4. (4) 1.51.5
    Show answer

    Answer: (2) 11

    1. P(A)+P(Aˉ)=1P(A)+P(\bar{A})=1

    2. x3+23=1\dfrac{x}{3}+\dfrac{2}{3}=1

    3. x3=13⇒x=1\dfrac{x}{3}=\dfrac{1}{3}\Rightarrow x=1

    Question 1 of 5

  2. The mean of 100100 observations is 4040 and their standard deviation is 33. The sum of squares of all observations is

    1. (1) 4000040000
    2. (2) 160900160900
    3. (3) 160000160000
    4. (4) 3000030000
    Show answer

    Answer: (2) 160900160900

    1. σ2=∑x2n−(xˉ)2\sigma^2=\dfrac{\sum x^2}{n}-\left(\bar{x}\right)^2

    2. 9=∑x2100−16009=\dfrac{\sum x^2}{100}-1600

    3. ∑x2100=1609\dfrac{\sum x^2}{100}=1609

    4. ∑x2=160900\sum x^2=160900

    Question 2 of 5

  3. Which of the following is incorrect?

    1. (1) P(A)>1P(A) > 1
    2. (2) 0≤P(A)≤10 \leq P(A) \leq 1
    3. (3) P(ϕ)=0P(\phi) = 0
    4. (4) P(A)+P(Aˉ)=1P(A) + P(\bar{A}) = 1
    Show answer

    Answer: (1) P(A)>1P(A) > 1

    1. Probability of any event always lies between 00 and 11.

    2. So P(A)>1P(A)>1 is impossible.

    Question 3 of 5

  4. The probability a red marble selected at random from a jar containing pp red, qq blue and r green marbles is

    1. (1) qp+q+r\dfrac{q}{p+q+r}
    2. (2) pp+q+r\dfrac{p}{p+q+r}
    3. (3) p+qp+q+r\dfrac{p+q}{p+q+r}
    4. (4) p+rp+q+r\dfrac{p+r}{p+q+r}
    Show answer

    Answer: (2) pp+q+r\dfrac{p}{p+q+r}

    1. Total marbles =p+q+r=p+q+r; red marbles =p=p.

    2. P(red)=pp+q+rP(\text{red})=\dfrac{p}{p+q+r}

    Question 4 of 5

  5. The range of the data 8,8,8,8,8...88, 8, 8, 8, 8. . . 8 is

    1. (1) 00
    2. (2) 11
    3. (3) 88
    4. (4) 33
    Show answer

    Answer: (1) 00

    1. Largest value =8= 8, smallest value =8= 8.

    2. Range =L−S=8−8=0= L - S = 8 - 8 = 0.

    Question 5 of 5
Chapter 8 formulas
  • Range

    R=L−S(L - largest value, S - smallest value)R = L - S \quad (L\text{ - largest value},\ S\text{ - smallest value})

  • Coefficient of range

    Coefficient of range=L−SL+S\text{Coefficient of range} = \frac{L-S}{L+S}

  • Variance

    σ2=∑i=1n(xi−xˉ)2n\sigma^{2} = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^{2}}{n}

  • Standard deviation

    σ=∑(xi−xˉ)2n\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^{2}}{n}}

  • Standard deviation (ungrouped data) - Direct method

    σ=∑xi2n−(∑xin)2\sigma = \sqrt{\frac{\sum x_i^{2}}{n} - \left(\frac{\sum x_i}{n}\right)^{2}}

  • Standard deviation (ungrouped data) - Mean method

    σ=∑di2n\sigma = \sqrt{\frac{\sum d_i^{2}}{n}}

  • Standard deviation (ungrouped data) - Assumed mean method

    σ=∑di2n−(∑din)2\sigma = \sqrt{\frac{\sum d_i^{2}}{n} - \left(\frac{\sum d_i}{n}\right)^{2}}

  • Standard deviation (ungrouped data) - Step deviation method

    σ=c×∑di2n−(∑din)2\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

    σ=n2−112\sigma = \sqrt{\frac{n^{2}-1}{12}}

  • Standard deviation (grouped data) - Mean method

    σ=∑fidi2N\sigma = \sqrt{\frac{\sum f_i d_i^{2}}{N}}

  • Standard deviation (grouped data) - Assumed mean method

    σ=∑fidi2N−(∑fidiN)2\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

    σ=C×∑fidi2N−(∑fidiN)2\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

    C.V=σxˉ×100%\text{C.V} = \frac{\sigma}{\bar{x}} \times 100\%

  • Consistency using C.V.

    Smaller C.V.⇒more consistent data;larger C.V.⇒less consistent data\text{Smaller C.V.} \Rightarrow \text{more consistent data};\quad \text{larger C.V.} \Rightarrow \text{less consistent data}

  • Mutually exclusive events

    A, B are mutually exclusive if A∩B=ϕA,\ B \text{ are mutually exclusive if } A \cap B = \phi

  • Probability of an event

    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

  • Sure and impossible events

    P(S)=1,P(ϕ)=0P(S) = 1,\qquad P(\phi) = 0

  • Range of probability

    0≤P(E)≤10 \le P(E) \le 1

  • Complement of an event

    P(Eˉ)=1−P(E)P(\bar{E}) = 1 - P(E)

  • Addition theorem (mutually exclusive events)

    P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

  • Probability of only A

    P(A∩Bˉ)=P(only A)=P(A)−P(A∩B)P(A \cap \bar{B}) = P(\text{only } A) = P(A) - P(A \cap B)

  • Probability of only B

    P(Aˉ∩B)=P(only B)=P(B)−P(A∩B)P(\bar{A} \cap B) = P(\text{only } B) = P(B) - P(A \cap B)

  • Addition theorem for any two events

    P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  • Addition theorem for any three events

    P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)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

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