Numerical Problems II: Number of Elements in a Set (Three Sets)

NP1: A research team interviewed 100 people and recorded their status regarding three conditions:

I = Idle

P = Poor

W = Willing to work

The results were as follows:

Category	Number of People
Idle (I)	28
Poor (P)	33
Willing to work (W)	14
Idle and Poor (I ∩ P)	12
Idle and Willing to work (I ∩ W)	9
Poor and Willing to work (P ∩ W)	6
Idle, Poor, and Willing (I ∩ P ∩ W)	5

1. Check whether the data are consistent.

2. Find how many people are either idle, poor, or willing to work, n(I ∪ P ∪ W).

Solution:

We use the inclusion-exclusion principle for 3 sets:

n(I ∪ P ∪ W) = n(I) + n(P) + n(W) − n(I ∩ P) − n(I ∩ W) − n(P ∩ W) + n(I ∩ P ∩ W) 

n(I ∪ P ∪ W) = 28 + 33 + 14 − 12 − 9 − 6 + 5

n(I ∪ P ∪ W) = 75 − 27 + 5=53 

n(I ∪ P ∪ W) =  53 

So, only 53 people are in I ∪ P ∪ W.

That means:

People not in any group, n(I' ∩ P' ∩ W')  = 100 − 53 = 47 

 ∵ The union count is ≤ 100 and all overlaps are consistent with the individual set sizes

∴ The data are consistent.

Also,

The number of people who are either idle, poor, or willing to work, n(I ∪ P ∪ W) = 53

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NP2: In an examination 53% students passed in Mathematics, 61% in Economics and 60% in Statistics. 24% passed in Mathematics as well as Economics, 35% in Mathematics and Statistics and 27% in Economics and Statistics. 5% passed in none of these subjects. How many students passed in all the 3 subjects? How many students failed in exactly one subject? How many passed in exactly one subject.

Solution:

Let, total number of students, n(Ω) = 100 students

Given:

Passed in Mathematics, n(M) = 53

Passed in Economics, n(E) = 61

Passed in Statistics, n(S) = 60

Passed in both Mathematics & Economics, n(M ∩ E) = 24

Passed in both Mathematics & Statistics, n(M ∩ S) = 35

Passed in both Economics & Statistics, n(E ∩ S) = 27

Passed in none of the subjects, n(M' ∪ E' ∪ S') = 27

Passed in at least one, n(M ∪ E ∪ S) = 100 − n(M' ∪ E' ∪ S') = 100 − 5 = 95 

To find: 

i) Number of students passed in all three subjects, n(M ∩ E ∩ S) 

ii) Number of students who passed in exactly one subject

iii) Number of students who failed in exactly one subject

Using the Union Formula for Three Sets:

n(M ∪ E ∪ S) = n(M) + n(E) + n(S)− n(M ∩ E) − n(M ∩ S) − n(E ∩ S) + n(M ∩ E ∩ S) 

95 = 53 + 61 + 60 − 24 − 35 − 27 + n(M ∩ E ∩ S) 

⇒95 = 174 − 86 +  

⇒95 = 88 + n(M ∩ E ∩ S)

⇒n(M ∩ E ∩ S) = 7​ 

∴ Students who passed in all three subjects = 7

Next,

Number of students who passed in exactly one subject

a) Only Mathematics: 

= n(M) − n(M ∩ E) − n(M ∩ S) + n(M ∩ E ∩ S) =53 − 24 − 35 + 7 = 1​ 

b) Only Economics:

= n(E) − n(M ∩ E) − n(E ∩ S) + n(M ∩ E ∩ S) = 61 − 24 − 27 + 7 = 17​ 

c) Only Statistics:

= n(S) − n(M ∩ S) − n(E ∩ S) + n(M ∩ E ∩ S) = 60 − 35 − 27 + 7 = 5​ 

∴ Passed in exactly one subject = 1 + 17 + 5 = 23​ 

Next, 

Number of students who failed in exactly one subject, i.e., they passed in two subjects.

a) Passed in Mathematics & Economics only:

= n(M ∩ E) − n(M ∩ E ∩ S) = 24 − 7 = 17​ 

b) Passed in Mathematics & Statistics only:

= n(M ∩ S) − n(M ∩ E ∩ S) = 35 − 7 = 28​ 

c) Passed in Economics & Statistics only:

= n(E ∩ S) − n(M ∩ E ∩ S) = 27 − 7 = 20​ 

∴ Passed in exactly two subjects (i.e., failed in exactly one) = 17 + 28 + 20 = 65​ 

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NP3: In a survey of 300 companies, the number of companies using different media for advertisements are as follows:

N = 200, R = 100, H = 80, R∩H = 30, N∩R = 50, N∩H = 50, N∩R∩H = 20

Find the number of companies using media other than newspapers (N), radio (R) and handbills (H) for advertisements.

Solution:

Given:

n(Ω) = 300

n(N) = 200

n(R) = 100

n (H) = 80

n(R∩H) = 30

n(N∩R) = 50

n(N∩H) = 50

n(N∩R∩H) = 20

To find: The number of companies using none of these three media, n(Ω) − n(N ∪ R ∪ H)

Using the Union Formula for Three Sets:

n(N ∪ R ∪ H) = n(N) + n(R) + n(H) − n(N ∩ R) − n(N ∩ H) − n(R ∩ H) + n(N ∩ R ∩ H) 

n(N ∪ R ∪ H) = 200 + 100 + 80 − 50 − 50 − 30 + 20 

n(N ∪ R ∪ H) = 380 − 130 + 20 

n(N ∪ R ∪ H) = 270

Now, 

n(Ω) − n(N ∪ R ∪ H) = 300 − 270 = 30

∴ 30 companies use media other than Newspapers, Radio, or Handbills for advertisements.

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NP4: In a survey conducted of 1000 teachers in Udaipur, it was found that 48% preferred coffee (C), 54% tea (T) and 64% were smokers (S), 28% took coffee as well as tea, 30% took tea and also smoked, and 30% were smokers and coffee users, 6% did none of these.

Find the number 

(i) using tea, coffee and also smoking

(ii) took only coffee, and

(iii) smokers taking tea but not coffee.

Solution:

Given:

n(Ω) = 100

n(C) = 48% = 480

n(T) = 54% = 540

n(S) = 64% → 640

n(C ∩ T) = 28% → 280

n(T ∩ S) = 30% → 300

n(S ∩ C) = 30% → 300

None of these, n(C' ∪ T' ∪ S') = 6% → 60

To find:

i) Number of teachers who drink coffee, tea, and smoke, n(C ∩ T ∩ S)

ii) Number who took only coffee, n(C) − n(C ∩ T) − n(S ∩ C) +  n(C ∩ T ∩ S)

iii) Number of smokers taking tea but not coffee, n(T ∩ S) −  n(C ∩ T ∩ S)

Now,

n(C ∪ T ∪S) = n(Ω) − n(C' ∪ T' ∪ S') = 1000 − 60 = 940  

Using the Union Formula for Three Sets:

n(C ∪ T ∪S) = n(C) + n(T) + n(S) − n(C ∩ T) − n(T ∩ S) − n(C ∩ S) + n(C ∩ T ∩ S) 

940 = 480 + 540 + 640 − 280 − 300 − 300 + n(C ∩ T ∩ S) 

940 = 1660 − 880 + n(C ∩ T ∩ S) 

940 = 780 + n(C ∩ T ∩ S) 

n(C ∩ T ∩ S) = 940 − 780  

n(C ∩ T ∩ S) = 160​ 

∴ Teachers who use tea, coffee, and also smoke = 160

Next,

Number who took only coffee, n(C) − n(C ∩ T) − n(S ∩ C) +  n(C ∩ T ∩ S) = 480 − 280 − 300 + 160 = 60​ 

∴ Teachers who took only coffee = 60

Next,

Number of smokers taking tea but not coffee, n(T ∩ S) −  n(C ∩ T ∩ S) = 300 − 160 = 140​ 

∴ Smokers who took tea but not coffee = 140

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