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IMO Answer D

Probability of at least one vehicle = P(Bus)+P(Auto)+P(Taxi)-Overlaps
= P(Bus)+P(Auto)+P(Taxi)-P(person likes two vehicles)-P(person likes three vehicles)
= 55%+37%+25%-30%-6%
= 81%
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how many people like to travel by at least one of the three vehicles:

x=55+37+25-30-6--> x=81%

Answer -> D
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Instead of using percentages, assume the total unique people = 100.

How many people travel by at least 1?

Re-phrase the question as --- what is the: TOTAL No. of Unique People (100) - (No. of people who like NONE) = ?

Key Phrase: "30% of the ppl like to travel by AT LEAST 2 vehicles, while 6% of ppl like ALL 3"

Total No. of PPL who like AT LEAST 2 = (No. of ppl who like EXACTLY 2) + (No. of ppl who like EXACTLY 3)

30 = (Exactly 2) + 6

No. of ppl who like EXACTLY 2 = 24


Because in the 55%, 37%, and 25% there is over-counting, we need to remove the Over-Counting for people who like 2 and for people who like 3.


Total No. of Unique People = (A + B + C) - (No. of ppl who like EXACTLY 2) - 2 * (No. of Pl who like EXACTLY 3) + NONE

100 = (55 + 37 + 25) - (24) - 2 * (6) + NONE

100 = 117 - 24 - 12 + NONE

100 = 81 + NONE

NONE = 19% = the Percentage of PPL who like NONE

Answer: Total No. of Unique People (100%) - No. of People who like NONE (19%) = No. of People who like AT LEAST 1 = 81%

Answer D
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IMO Answer D

Probability of at least one vehicle = P(Bus)+P(Auto)+P(Taxi)-Overlaps
= P(Bus)+P(Auto)+P(Taxi)-P(person likes two vehicles)-P(person likes three vehicles)
= 55%+37%+25%-30%-6%
= 81%

Wrong formula. Probability of the union of three events A,B, and C is

P(AUBUC)=P(A)+P(B)+P(C)-P(AUB)-P(AUC)-P(BUC)+P(AUBUC)
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B=people that like to travel by Bus
A=people that like travelling by auto
T=people that like travelling by taxi

Given:
P(B)=.55
P(A)=.37
P(T)=.25
P((BUA)U(BUT)U(AUT))=.30
P(B,A,T)=.06

We want to find P(BUAUT)

P(BUAUT)=P(B)+P(A)+P(T)-P(BUA)-P(BUT)-P(AUT)+P(A,B,T)=P(B)+P(A)+P(T)-[P(BUA)+P(BUT)+P(AUT)]+P(A,B,T)

In order to use this formula we need to get P(BUA)+P(BUT)+P(AUT), b/c the rest we already have.
(The mistake that I think some people made, including myself, was that P(BUA)+P(BUT)+P(AUT)=.3, but it is not)

Notice that in P((BUA)U(BUT)U(AUT)), P(A,B,T) is only counted once. While in P(BUA)+P(BUT)+P(AUT) it is counted three times. So we only need to add P(A,B,T) two more times to P((BUA)U(BUT)U(AUT)) to get P(BUA)+P(BUT)+P(AUT).

That is, P(BUA)+P(BUT)+P(AUT)=2P(A,B,T)+P((BUA)U(BUT)U(AUT))=2(.06)+.3=.42

Therefore ,

P(BUAUT)=P(B)+P(A)+P(T)-[P(BUA)+P(BUT)+P(AUT)]+P(A,B,T)=.55+.37+.25-.42+.06=1.17-.42+.06=.81

Final Answer: D
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