AIEEE 2002 Math Solved Paper

Given P(A)=‌

1

2

,P(B)=‌

1

3

,P(C)=‌

1

4

So, P(A)=‌

1

2

(Probability that the problem can't be solve by A)
P(B)=‌

2

3

(Probability that the problem can't be solve by B)
and P(C)=‌

3

4

(Probability that the problem can't be solve by C )
Now the probability that the problem is solved by any one student of A,B and C=1 - the probability that the problem is solved by none of the students of A,B and C
P(A∪B∪C)=1−P(A)P(B)P(C)
=1−‌

1

2

×‌

2

3

×‌

3

4

=‌

3

4