A sample of 100 people was surveyed for the use of computers, cell phones and credit cards. 31 people had credit cards, 61 had computers and 28 had cell phones. If 11 people had none of the three and 32 people had more than one, how many had all the three?
Solution: This is a question based on Set Theory.
Total = 100
None = 11
Net Total = 100 - 11
P(A) = Credit Cards = 31
P(B) = Computers = 61
P(C) = Cell Phones = 28
Using the formula, P(S) = P(A) + P(B) + P(C) - P(A?B) – P(B?C) – P(C?A) + P(A?B?C)
P (A?B) + P(B?C) + P(C?A) = 32 (More than One)
Plugging in values, we get,
100 – 11 = 31 + 61 + 28 – (P (A?B) + P(B?C) + P(C?A)) + P(A?B?C)
89 = 120 – (32 + P(A?B?C))
P(A?B?C) = 1
Correct Answer: A
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