If A & B are two events then `P{(AnnbarB)uu(barAnnB)} =`

If A o* B are two events then P{(A nnbar(B))uu(bar(A)nn B)}=(i)P(A uu B)-P(A nn B)(ii)P(A uu B)+P(A nn B)( iii) P(A)+P(B)(iv)P(A)+P(B)+P(A nn B)

If A and B are any two events then P(barAnnB) =

If A and B are two events. Then, 1+P(AnnB)-P(B)-P(A) is equal to a) P(barAnnbarB) b) P(AnnbarB) c) P(barAnnB) d) P(AuuB)

If A and B are two events such that P(A uu B)=P(A nn B), then the incorrect statement amongst the followving statements is :

If A and B are any two events having P(AnnB)=1/2 an P(A)=2/3 , then the probability of AnnbarB is :

If A and B are two events such that P(A uu B) + P(A nn B) = (7)/(8) and P(A) = 2P(B) , then P(A) =

If A and B are two events such that P(A uu B) + P(A nn B) = (7)/(8) and P(A) = 2P(B) , then P(A) =

If A and B are two events such that P(A uu B) = P(A nn B) then the true realtion is