AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

It is assumed the events.
E1=6 occurs
E2=6 does not occur
A= mean reports that it is 6
We have,
P(

A

E1

)=

2

3

P(

A

E2

)=

1

3

The expression for probability that it is actually 6 is,
p(

E1

A

)=

1 6 × 2 3

( 1 6 × 2 3 )+( 5 6 × 1 3 )

=

2

7

The probability that it is not actually 6 is.
P=1−

2

7

=

5

7

The probability that it is actually 5 is,
P=

1

5

×

5

7

=

1

7