This is a question from one of MGMAT tests.
In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
I just want to understand why this question cannot be solved by using equation
AUBUC = A + B +C - AintB - BintC - CintA + AintBintC
The above equation will result is 22 however the question asks for exactly two classes.
P(AnB) + P(AnC) + P(BnC) - 3P(AnBnC)
= 22- 3(3) = 22 - 9 = 13
What is the OA?
My Apologies, that my upper post has tyoe. I don't know how it happen
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
68 = 25+ 25+ 34 - P(A n B) – P(A n C) – P(B n C) + 3
P(A n B) + P(A n C) + P(B n C) = 19
for exactly two persons we need to find out.
P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
= 19 - 3(3) = 10
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