AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

As the probability of occurrence of event A is,
P(A)=

2

5

Then, the probability of non-occurrence of event A is,
P(A)′=1−P(A)
=1−

2

5

=

3

5

As the probability of non-occurrence of event B is,
P(B)′=

3

10

Then, the probability of occurrence of event B is,
P(B)=1−P(B)′
=1−

3

10

=

7

10

The probability that only one of the two events occur is,
P(A)P(B′)+P(A′)P(B) =

2

5

(

3

10

)+35(

7

10

)
=

6

50

+

21

50

=

27

50