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Samacheer Kalvi 10th Maths Exercise 8.1 Solutions

10th Maths Exercise 8.1 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 16 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance16 answers · full steps below
  1. Q1 (i)Range =62=62, Coefficient of range =0.33=0.33
  2. Q1 (ii)Range =47.8=47.8, Coefficient of range =0.64=0.64
  3. Q2Largest value =50.2=50.2
  4. Q3Range =250=250 (Coefficient of range =0.238=0.238)
  5. Q4Standard deviation ≈2.34\approx 2.34 pages
  6. Q5Variance =222.22=222.22, Standard deviation ≈14.91\approx 14.91
  7. Q6The bell strikes 156156 times in a day; standard deviation σ=14312≃3.45\sigma=\sqrt{\dfrac{143}{12}}\simeq 3.45
  8. Q7Standard deviation ≈6.06\approx 6.06
  9. Q8New standard deviation =4.5=4.5
  10. Q9New standard deviation =1.2=1.2, New variance =1.44=1.44
  11. Q10Standard deviation ≈7.76\approx 7.76 mm
  12. Q11Standard deviation ≈14.61\approx 14.61
  13. Q12Standard deviation ≈5.99\approx 5.99 cm
  14. Q13Standard deviation ≈1.24\approx 1.24 sec
  15. Q14Correct mean =60.5=60.5, Correct standard deviation ≈14.61\approx 14.61
  16. Q15The remaining two observations are 66 and 88

Q1. Find the range and coefficient of range of the following data.

(i)

63,89,98,125,79,108,117,6863, 89, 98, 125, 79, 108, 117, 68

  1. Largest value L=125L=125, smallest value S=63S=63.

  2. Range R=L−S=125−63=62R = L-S = 125-63 = 62.

  3. Coefficient of range =L−SL+S=62188= \dfrac{L-S}{L+S} = \dfrac{62}{188}.

  4. =0.3297≈0.33= 0.3297 \approx 0.33.

Answer

Range =62=62, Coefficient of range =0.33=0.33

(ii)

43.5,13.6,18.9,38.4,61.4,29.843.5, 13.6, 18.9, 38.4, 61.4, 29.8

  1. Largest value L=61.4L=61.4, smallest value S=13.6S=13.6.

  2. Range R=61.4−13.6=47.8R = 61.4-13.6 = 47.8.

  3. Coefficient of range =61.4−13.661.4+13.6=47.875= \dfrac{61.4-13.6}{61.4+13.6} = \dfrac{47.8}{75}.

  4. =0.6373≈0.64= 0.6373 \approx 0.64.

Answer

Range =47.8=47.8, Coefficient of range =0.64=0.64

Q2. If the range and the smallest value of a set of data …

If the range and the smallest value of a set of data are 36.836.8 and 13.413.4 respectively, then find the largest value.

  1. Range R=L−SR = L - S.

  2. 36.8=L−13.436.8 = L - 13.4

  3. L=36.8+13.4=50.2L = 36.8 + 13.4 = 50.2.

Answer

Largest value =50.2=50.2

Q3. Calculate the range of the following data. …

Calculate the range of the following data.

Income 400-450 450-500 500-550 550-600 600-650
Number of workers 8 12 30 21 6

  1. For a continuous distribution, Range == upper limit of the highest class −- lower limit of the lowest class.

  2. L=650L = 650, S=400S = 400.

  3. Range R=650−400=250R = 650-400 = 250.

  4. Coefficient of range =650−400650+400=2501050=0.238=\dfrac{650-400}{650+400}=\dfrac{250}{1050}=0.238.

Answer

Range =250=250 (Coefficient of range =0.238=0.238)

Q4. A teacher asked the students to complete 6060 pages of a record …

A teacher asked the students to complete 6060 pages of a record note book. Eight students have completed only 32,35,37,30,33,36,3532, 35, 37, 30, 33, 36, 35 and 3737 pages. Find the standard deviation of the pages completed by them.

Show steps (7)

  1. Let the assumed mean A=35A = 35, n=8n = 8.

  2. di=xi−35d_i = x_i - 35 : −3,  0,  2,  −5,  −2,  1,  0,  2-3,\;0,\;2,\;-5,\;-2,\;1,\;0,\;2

  3. Σdi=−5\Sigma d_i = -5

  4. di2d_i^2 : 9,0,4,25,4,1,0,49,0,4,25,4,1,0,4 so Σdi2=47\Sigma d_i^2 = 47.

  5. σ=Σdi2n−(Σdin)2\sigma=\sqrt{\dfrac{\Sigma d_i^{2}}{n}-\left(\dfrac{\Sigma d_i}{n}\right)^{2}}

  6. =478−(−58)2=5.875−0.3906=\sqrt{\dfrac{47}{8}-\left(\dfrac{-5}{8}\right)^{2}}=\sqrt{5.875-0.3906}

  7. =5.4844≈2.34=\sqrt{5.4844}\approx 2.34

Answer

Standard deviation ≈2.34\approx 2.34 pages

Q5. Find the variance and standard deviation of the wages of 99 workers …

Find the variance and standard deviation of the wages of 99 workers given below: ₹310, ₹290, ₹320, ₹280, ₹300, ₹290, ₹320, ₹310, ₹280.

Show steps (6)

  1. Σxi=310+290+320+280+300+290+320+310+280=2700\Sigma x_i = 310+290+320+280+300+290+320+310+280 = 2700, n=9n = 9.

  2. Mean xˉ=27009=300\bar x = \dfrac{2700}{9}=300.

  3. di=xi−300d_i = x_i-300 : 10,−10,20,−20,0,−10,20,10,−2010,-10,20,-20,0,-10,20,10,-20

  4. Σdi2=100+100+400+400+0+100+400+100+400=2000\Sigma d_i^{2}=100+100+400+400+0+100+400+100+400=2000

  5. Variance σ2=Σdi2n=20009=222.22\sigma^{2}=\dfrac{\Sigma d_i^{2}}{n}=\dfrac{2000}{9}=222.22

  6. σ=222.22≈14.91\sigma=\sqrt{222.22}\approx 14.91

Answer

Variance =222.22=222.22, Standard deviation ≈14.91\approx 14.91

Q6. A wall clock strikes the bell once at 11 o' clock, 22 …

A wall clock strikes the bell once at 11 o' clock, 22 times at 22 o' clock, 33 times at 33 o' clock and so on. How many times will it strike in a particular day. Find the standard deviation of the number of strikes the bell make a day.

Show steps (6)

  1. In 1212 hours the bell strikes 1+2+3+⋯+12=12×132=781+2+3+\cdots+12=\dfrac{12\times 13}{2}=78 times, so in a day (two rounds of 1212 hours) it strikes 2×78=1562\times 78=156 times.

  2. Hence the hourly numbers of strikes in a day are 1,2,3,…,12,1,2,3,…,121,2,3,\ldots,12,1,2,3,\ldots,12, so n=24n=24 and Σx=156\Sigma x=156.

  3. Mean xˉ=Σxn=15624=6.5\bar{x}=\dfrac{\Sigma x}{n}=\dfrac{156}{24}=6.5.

  4. Σx2=2(12+22+⋯+122)=2×12×13×256=2×650=1300\Sigma x^{2}=2\left(1^{2}+2^{2}+\cdots+12^{2}\right)=2\times\dfrac{12\times 13\times 25}{6}=2\times 650=1300.

  5. σ=Σx2n−(Σxn)2=130024−(6.5)2=54.1667−42.25=11.9167\sigma=\sqrt{\dfrac{\Sigma x^{2}}{n}-\left(\dfrac{\Sigma x}{n}\right)^{2}}=\sqrt{\dfrac{1300}{24}-(6.5)^{2}}=\sqrt{54.1667-42.25}=\sqrt{11.9167}.

  6. σ≃3.45\sigma\simeq 3.45 (this equals 122−112\sqrt{\dfrac{12^{2}-1}{12}}, since repeating the values 11 to 1212 twice does not change the spread).

Answer

The bell strikes 156156 times in a day; standard deviation σ=14312≃3.45\sigma=\sqrt{\dfrac{143}{12}}\simeq 3.45

Q7. Find the standard deviation of first 2121 natural numbers.

  1. For the first nn natural numbers, σ=n2−112\sigma=\sqrt{\dfrac{n^{2}-1}{12}}.

  2. Here n=21n = 21.

  3. σ=441−112=44012\sigma=\sqrt{\dfrac{441-1}{12}}=\sqrt{\dfrac{440}{12}}

  4. =36.67≈6.06=\sqrt{36.67}\approx 6.06

Answer

Standard deviation ≈6.06\approx 6.06

Q8. If the standard deviation of a data is 4.54.5 and if each …

If the standard deviation of a data is 4.54.5 and if each value of the data is decreased by 55, then find the new standard deviation.

  1. Subtracting a fixed constant from every value does not change the spread of the data.

  2. So the standard deviation remains the same.

Answer

New standard deviation =4.5=4.5

Q9. If the standard deviation of a data is 3.63.6 and each value …

If the standard deviation of a data is 3.63.6 and each value of the data is divided by 33, then find the new variance and new standard deviation.

  1. When each value is divided by kk, the standard deviation also gets divided by kk.

  2. New σ=3.63=1.2\sigma=\dfrac{3.6}{3}=1.2

  3. New variance =(1.2)2=1.44=(1.2)^{2}=1.44

Answer

New standard deviation =1.2=1.2, New variance =1.44=1.44

Q10. The rainfall recorded in various places of five districts in a week …

The rainfall recorded in various places of five districts in a week are given below. Find its standard deviation.

Rainfall (in mm) 45 50 55 60 65 70
Number of places 5 13 4 9 5 4
Show steps (7)

  1. Assumed mean A=60A = 60, c=5c = 5, N=5+13+4+9+5+4=40N = 5+13+4+9+5+4 = 40.

  2. di=xi−605d_i=\dfrac{x_i-60}{5} : −3,−2,−1,0,1,2-3,-2,-1,0,1,2

  3. fidif_id_i : −15,−26,−4,0,5,8⇒Σfidi=−32-15,-26,-4,0,5,8 \Rightarrow \Sigma f_id_i=-32

  4. fidi2f_id_i^{2} : 45,52,4,0,5,16⇒Σfidi2=12245,52,4,0,5,16 \Rightarrow \Sigma f_id_i^{2}=122

  5. σ=c×Σfidi2N−(ΣfidiN)2\sigma=c\times\sqrt{\dfrac{\Sigma f_id_i^{2}}{N}-\left(\dfrac{\Sigma f_id_i}{N}\right)^{2}}

  6. =5×12240−(−3240)2=53.05−0.64=5\times\sqrt{\dfrac{122}{40}-\left(\dfrac{-32}{40}\right)^{2}}=5\sqrt{3.05-0.64}

  7. =52.41=5×1.552≈7.76=5\sqrt{2.41}=5\times 1.552\approx 7.76

Answer

Standard deviation ≈7.76\approx 7.76 mm

Q11. In a study about viral fever, the number of people affected in …

In a study about viral fever, the number of people affected in a town were noted as Find its standard deviation.

Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Number of people affected 3 5 16 18 12 7 4
Show steps (7)

  1. Mid values xix_i : 5,15,25,35,45,55,655,15,25,35,45,55,65; A=35A = 35, c=10c = 10, N=65N = 65.

  2. di=xi−3510d_i=\dfrac{x_i-35}{10} : −3,−2,−1,0,1,2,3-3,-2,-1,0,1,2,3

  3. fidif_id_i : −9,−10,−16,0,12,14,12⇒Σfidi=3-9,-10,-16,0,12,14,12 \Rightarrow \Sigma f_id_i=3

  4. fidi2f_id_i^{2} : 27,20,16,0,12,28,36⇒Σfidi2=13927,20,16,0,12,28,36 \Rightarrow \Sigma f_id_i^{2}=139

  5. σ=c×Σfidi2N−(ΣfidiN)2\sigma=c\times\sqrt{\dfrac{\Sigma f_id_i^{2}}{N}-\left(\dfrac{\Sigma f_id_i}{N}\right)^{2}}

  6. =10×13965−(365)2=102.1385−0.0021=10\times\sqrt{\dfrac{139}{65}-\left(\dfrac{3}{65}\right)^{2}}=10\sqrt{2.1385-0.0021}

  7. =10×1.4616≈14.61=10\times 1.4616\approx 14.61

Answer

Standard deviation ≈14.61\approx 14.61

Q12. The measurements of the diameters (in cms) of the plates prepared in …

The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.

Diameter (cm) 21-24 25-28 29-32 33-36 37-40 41-44
Number of plates 15 18 20 16 8 7
Show steps (7)

  1. The classes are inclusive, so take true boundaries 20.5 ⁣− ⁣24.5,  24.5 ⁣− ⁣28.5,…20.5\!-\!24.5,\;24.5\!-\!28.5,\ldots

  2. Mid values xix_i : 22.5, 26.5, 30.5, 34.5, 38.5, 42.522.5,\,26.5,\,30.5,\,34.5,\,38.5,\,42.5; A=34.5A=34.5, c=4c=4, N=84N=84.

  3. di=xi−34.54d_i=\dfrac{x_i-34.5}{4} : −3,−2,−1,0,1,2-3,-2,-1,0,1,2

  4. fidif_id_i : −45,−36,−20,0,8,14⇒Σfidi=−79-45,-36,-20,0,8,14 \Rightarrow \Sigma f_id_i=-79

  5. fidi2f_id_i^{2} : 135,72,20,0,8,28⇒Σfidi2=263135,72,20,0,8,28 \Rightarrow \Sigma f_id_i^{2}=263

  6. σ=4×26384−(−7984)2=43.131−0.885\sigma=4\times\sqrt{\dfrac{263}{84}-\left(\dfrac{-79}{84}\right)^{2}}=4\sqrt{3.131-0.885}

  7. =42.246=4×1.4988≈5.99=4\sqrt{2.246}=4\times 1.4988\approx 5.99

Answer

Standard deviation ≈5.99\approx 5.99 cm

Q13. The time taken by 50 students to complete a 100 meter race …

The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Time taken (sec) 8.5-9.5 9.5-10.5 10.5-11.5 11.5-12.5 12.5-13.5
Number of students 6 8 17 10 9
Show steps (6)

  1. Mid values xix_i : 9,10,11,12,139,10,11,12,13; A=11A = 11, c=1c = 1, N=50N = 50.

  2. di=xi−111d_i=\dfrac{x_i-11}{1} : −2,−1,0,1,2-2,-1,0,1,2

  3. fidif_id_i : −12,−8,0,10,18⇒Σfidi=8-12,-8,0,10,18 \Rightarrow \Sigma f_id_i=8

  4. fidi2f_id_i^{2} : 24,8,0,10,36⇒Σfidi2=7824,8,0,10,36 \Rightarrow \Sigma f_id_i^{2}=78

  5. σ=1×7850−(850)2=1.56−0.0256\sigma=1\times\sqrt{\dfrac{78}{50}-\left(\dfrac{8}{50}\right)^{2}}=\sqrt{1.56-0.0256}

  6. =1.5344≈1.24=\sqrt{1.5344}\approx 1.24

Answer

Standard deviation ≈1.24\approx 1.24 sec

Q14. For a group of 100100 candidates the mean and standard deviation of …

For a group of 100100 candidates the mean and standard deviation of their marks were found to be 6060 and 1515 respectively. Later on it was found that the scores 4545 and 7272 were wrongly entered as 4040 and 2727. Find the correct mean and standard deviation.

Show steps (8)

  1. n=100n=100, xˉ=60⇒Σx=100×60=6000\bar x=60 \Rightarrow \Sigma x = 100\times 60 = 6000.

  2. Correct total =6000−40−27+45+72=6050=6000-40-27+45+72 = 6050.

  3. Correct mean =6050100=60.5=\dfrac{6050}{100}=60.5.

  4. σ=15⇒225=Σx2100−602\sigma=15 \Rightarrow 225=\dfrac{\Sigma x^{2}}{100}-60^{2}

  5. Σx2=100(225+3600)=382500\Sigma x^{2}=100(225+3600)=382500 (incorrect value).

  6. Correct Σx2=382500−402−272+452+722=387380\Sigma x^{2}=382500-40^{2}-27^{2}+45^{2}+72^{2}=387380.

  7. Correct σ=387380100−(60.5)2=3873.8−3660.25\sigma=\sqrt{\dfrac{387380}{100}-(60.5)^{2}}=\sqrt{3873.8-3660.25}

  8. =213.55≈14.61=\sqrt{213.55}\approx 14.61

Answer

Correct mean =60.5=60.5, Correct standard deviation ≈14.61\approx 14.61

Q15. The mean and variance of seven observations are 88 and 1616 respectively. …

The mean and variance of seven observations are 88 and 1616 respectively. If five of these are 2,4,10,122, 4, 10, 12 and 1414, then find the remaining two observations.

Show steps (9)

  1. n=7n=7, mean =8⇒Σx=56=8 \Rightarrow \Sigma x = 56.

  2. Sum of the five given values =2+4+10+12+14=42=2+4+10+12+14=42.

  3. Let the other two be aa and bb: a+b=56−42=14a+b = 56-42 = 14 … (1)

  4. Variance =16⇒Σx27−82=16⇒Σx2=7×80=560=16 \Rightarrow \dfrac{\Sigma x^{2}}{7}-8^{2}=16 \Rightarrow \Sigma x^{2}=7\times 80=560.

  5. Sum of squares of the five values =4+16+100+144+196=460=4+16+100+144+196=460.

  6. So a2+b2=560−460=100a^{2}+b^{2}=560-460=100 … (2)

  7. (a+b)2=a2+b2+2ab⇒196=100+2ab⇒ab=48(a+b)^{2}=a^{2}+b^{2}+2ab \Rightarrow 196 = 100+2ab \Rightarrow ab = 48

  8. (a−b)2=196−4(48)=4⇒a−b=2(a-b)^{2}=196-4(48)=4 \Rightarrow a-b = 2

  9. Solving a+b=14a+b=14, a−b=2a-b=2 gives a=8a=8, b=6b=6.

Answer

The remaining two observations are 66 and 88

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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