Bullet wrote:

sinharavi wrote:

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?

A. 13

B. 10

C. 9

D. 8

E. 7

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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