Correct Answer - A
Given , `P(bar(A cup B)) =(1)/(6) , P(A cap B) =(1)/(4) ,P(bar(A)) =(1)/(4)`
`:. P(A cup B )=1-P(bar(A cup B)) =1-(1)/(6)=(5)/(6)`
` and P(A) =1-P(bar(A)) =1-(1)/(4)=(3)/(4)`
`:. P(A cup B ) =P(A) + P(B) -P( cap B ) `
`(5)/(6) =(3)/(4) +P(B)-(1)/(4)`
`implies P(B)=(1)/(3) implies ` A and are not equally likely
`P( A cap B ) =P(A) * P(B) =(1)/(4)`
So , events are independent .