Sur­veys often report how many peo­ple like one thing, how many like another, and how many like both. Sets and Venn dia­grams turn such reports into exact counts: how many like nei­ther, how many like only one, how many like all three. The skill is in trans­lat­ing the words of the prob­lem into the right set, so this les­son starts with a dic­tio­nary of phrases and then solves prob­lems with two and three sets.

From words to sets

Phrase in the prob­lem Set
at least one of AA, BB A∪BA \cup B
either AA or BB A∪BA \cup B
both AA and BB A∩BA \cap B
only AA A−BA - B
exactly one of AA, BB (A−B)∪(B−A)=A Δ B(A - B) \cup (B - A) = A \,\Delta\, B
nei­ther AA nor BB A′∩B′=(A∪B)′A' \cap B' = (A \cup B)'

"Each per­son likes at least one" means nobody is out­side both cir­cles, so n(A∪B)n(A \cup B) is the whole group.

Two sets

Exam­ple 1

In a group of 70 peo­ple, 37 like cof­fee, 52 like tea, and each per­son likes at least one of the two drinks. How many like both cof­fee and tea?

Let CC and TT be the sets of peo­ple who like cof­fee and tea. Then n(C)=37n(C) = 37, n(T)=52n(T) = 52 and n(C∪T)=70n(C \cup T) = 70. By the for­mula,

n(C∪T)=n(C)+n(T)−n(C∩T)  ⟹  70=37+52−n(C∩T)n(C \cup T) = n(C) + n(T) - n(C \cap T) \implies 70 = 37 + 52 - n(C \cap T)

So n(C∩T)=89−70=19n(C \cap T) = 89 - 70 = 19. Nine­teen peo­ple like both.

Going fur­ther: only cof­fee =37−19=18= 37 - 19 = 18, and only tea =52−19=33= 52 - 19 = 33. Check: 18+19+33=7018 + 19 + 33 = 70.

Exam­ple 2

In a class of 35 stu­dents, 24 like to play cricket and 16 like to play foot­ball. Each stu­dent likes to play at least one of the two games. How many like to play both?

n(C∪F)=35n(C \cup F) = 35, so 35=24+16−n(C∩F)35 = 24 + 16 - n(C \cap F), giv­ing n(C∩F)=40−35=5n(C \cap F) = 40 - 35 = 5.

Exam­ple 3

In a class of 60 stu­dents, 25 learn music, 30 learn dance and 10 learn both. How many learn nei­ther?

n(M∪D)=25+30−10=45n(M \cup D) = 25 + 30 - 10 = 45. The num­ber who learn nei­ther is n((M∪D)′)=60−45=15n\left((M \cup D)'\right) = 60 - 45 = 15.

Exam­ple 4

A sur­vey shows that 63% of the peo­ple in a city like tea and 76% like cof­fee. If x%x\% like both tea and cof­fee, find the pos­si­ble val­ues of xx.

Think of the city as 100 peo­ple, so n(U)=100n(U) = 100, n(T)=63n(T) = 63, n(C)=76n(C) = 76 and n(T∩C)=xn(T \cap C) = x.

Upper limit. T∩CT \cap C is a sub­set of TT and of CC, so x≤min⁡{63,76}=63x \le \min\{63, 76\} = 63.

Lower limit. n(T∪C)n(T \cup C) can­not be more than the whole pop­u­la­tion: 63+76−x≤10063 + 76 - x \le 100, so 139−x≤100139 - x \le 100 and x≥39x \ge 39.

So 39≤x≤6339 \le x \le 63. At least 39% and at most 63% of the peo­ple like both drinks.

Three sets

With three sets, fill in the Venn dia­gram from the inside out: first the cen­tre (all three), then the "two only" regions, then the "only one" regions. Each region is found by sub­tract­ing what is already known.

Exam­ple 5

Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played bas­ket­ball. Of these, 64 played both bas­ket­ball and hockey, 80 played cricket and bas­ket­ball, 40 played cricket and hockey, and 24 played all three games. How many boys did not play any game?

Let CC, HH, BB be the three sets.

By the for­mula:

n(C∪H∪B)=224+240+336−64−80−40+24=640n(C \cup H \cup B) = 224 + 240 + 336 - 64 - 80 - 40 + 24 = 640

So the num­ber who played no game is 800−640=160800 - 640 = 160.

Region by region, which also shows how many played only one game:

Region Work­ing Boys
all three given 24
cricket and bas­ket­ball only 80−2480 - 24 56
bas­ket­ball and hockey only 64−2464 - 24 40
cricket and hockey only 40−2440 - 24 16
cricket only 224−(56+24+16)224 - (56 + 24 + 16) 128
hockey only 240−(16+24+40)240 - (16 + 24 + 40) 160
bas­ket­ball only 336−(56+24+40)336 - (56 + 24 + 40) 216

The seven regions add up to 24+56+40+16+128+160+216=64024 + 56 + 40 + 16 + 128 + 160 + 216 = 640, as before, and 800−640=160800 - 640 = 160 boys played no game.

Exam­ple 6

In a sur­vey of 100 peo­ple about three news­pa­pers AA, BB and CC: 40 read AA, 35 read BB, 30 read CC, 12 read AA and BB, 10 read BB and CC, 8 read AA and CC, and 5 read all three. Find how many read at least one paper, none, exactly one, and exactly two.

Here n(A)+n(B)+n(C)=105n(A) + n(B) + n(C) = 105 and the pair­wise inter­sec­tions add up to 30.

At least one: 105−30+5=80105 - 30 + 5 = 80. None: 100−80=20100 - 80 = 20.

Exactly two: 30−3×5=1530 - 3 \times 5 = 15. Exactly one: 105−2×30+3×5=60105 - 2 \times 30 + 3 \times 5 = 60.

Check: 60+15+5=8060 + 15 + 5 = 80.

Prac­tice

  1. In a group of 50 peo­ple, 30 speak Tel­ugu and 35 speak Eng­lish, and every­one speaks at least one of the two. How many speak both?
  2. In a group of 65 peo­ple, each likes cricket or ten­nis or both. If 40 like cricket and 10 like both, how many like ten­nis? How many like ten­nis only?
  3. In a class of 120 stu­dents, 70 study Math­e­mat­ics, 60 study Sci­ence and 30 study both. How many study nei­ther?
  4. In a town, 55% of the peo­ple read an Eng­lish paper and 70% read a Tel­ugu paper. If x%x\% read both, find the range of pos­si­ble val­ues of xx.
  5. Of 200 peo­ple, 120 read paper PP, 90 read paper QQ, and 50 read nei­ther. How many read both?
  6. Of 100 stu­dents, 40 play foot­ball, 45 cricket and 35 hockey; 15 play foot­ball and cricket, 12 cricket and hockey, 10 foot­ball and hockey, and 5 play all three. How many play at least one game? How many play none?
  7. For ques­tion 6, how many play exactly one game, and how many play exactly two?
  8. For ques­tion 6, how many play cricket only?
  9. In a colony of 5000 fam­i­lies, 50% watch chan­nel P, 30% watch Q and 20% watch R; 10% watch P and Q, 6% Q and R, 8% P and R, and 4% watch all three. How many fam­i­lies watch P only? How many watch none of the three?

Answers

Show answers
  1. 30+35−50=1530 + 35 - 50 = 15
  2. n(T)=65−40+10=35n(T) = 65 - 40 + 10 = 35; ten­nis only =35−10=25= 35 - 10 = 25.
  3. n(M∪S)=100n(M \cup S) = 100, so nei­ther =120−100=20= 120 - 100 = 20.
  4. Upper limit 55; lower limit from 125−x≤100125 - x \le 100, so 25≤x≤5525 \le x \le 55.
  5. n(P∪Q)=150n(P \cup Q) = 150, so both =120+90−150=60= 120 + 90 - 150 = 60.
  6. 120−37+5=88120 - 37 + 5 = 88 play at least one; 100−88=12100 - 88 = 12 play none.
  7. Exactly two: 37−15=2237 - 15 = 22. Exactly one: 120−74+15=61120 - 74 + 15 = 61. (Check: 61+22+5=8861 + 22 + 5 = 88.)
  8. Cricket and foot­ball only =10= 10, cricket and hockey only =7= 7, so cricket only =45−(10+7+5)=23= 45 - (10 + 7 + 5) = 23.
  9. P only =50−10−8+4=36%= 50 - 10 - 8 + 4 = 36\%, that is 1800 fam­i­lies. At least one =100−24+4=80%= 100 - 24 + 4 = 80\%, so 20%, that is 1000 fam­i­lies, watch none.