If `E_(1)` and `E_(2)` are two events such that `P(E_(1))=1//4`, `P(E_(2)//E_(1))=1//2` and `P(E_(1)//E_(2))=1//4`, then

Options

A. then `E_(1)` and `E_(2)` are independent
B. `E_(1)` and `E_(2)` are exhaustive
C. `E_(2)` is twice as likely to occur as `E_(1)`
D. Probabilites of the events
`E_(1) nn E_(2)`, `E_(1)` and `E_(2)` are in `G.P.`

Answer

A, C, D

Explanation

`(a,c,d)` `P(E_(2)//E_(1))=(P(E_(1)nnE_(2)))/(P(E_(1)))`
`(1)/(2)=(P(E_(1)nnE_(2)))/(1//4)`
`impliesP(E_(1)nnE_(2))=(1)/(8)=P(E_(2))*P(E_(1)//E_(2))`
`=P(E_(2))*(1)/(4)`
`impliesP(E_(2))=(1)/(2)`
Since `P(E_(1) nnE_(2))=(1)/(8)=P(E_(1))*P(E_(1))*P(E_(2))`, events are independent
Also `P(E_(1)uuE_(2))=(1)/(2)+(1)/(4)-(1)/(8)=(5)/(8)`
`impliesE_(1)` and `E_(2)` are not exhaustive.