If `E_1, E_2`, are two events with `E_1 nn E_2 = phi` then `P(bar E_1 nn bar E_2) =`

If E_(1),E_(2) are two events with E_(1)capE_(2)=phi then show that P(E_(1)^(C)capE_(2)^(C))=P(E^(C))-P(E_(2))

Statement -1 : Let E_1, E_2, E_3 be three events such that P(E_1)+P(E_2)+P(E_3)=1, " then " E_1, E_2, E_3 are exhaustive events. Statement-2 if the events E_1, E_2 " and " E_3 be exhaustive events, then P(E_1cupE_2cupE_3)=1

Statement -1 : Let E_1, E_2, E_3 be three events such that P(E_1)+P(E_2)+P(E_3)=1, " then " E_1, E_2, E_3 are exhaustive events. Statement-2 if the events E_1, E_2 " and " E_3 be exhaustive events, then P(E_1cupE_2cupE_3)=1

If E_1 and E_2 are independent events associated with an experiment and P(E_1capE_2)=lambdaP(E_1)P(E_2) , then what is down the value of lambda .

Fill in the Blanks : If E_1 and E_2 are independent events, then P( E_1 nn E_2 ) =..

If E_(1), E_(2), E_(3) are independent events, such that P(E_(1)nnbar(E_(2))nnbar(E_(3)))=1/4, P(bar(E_(1))nnbar(E_(2))nnbar(E_(3)))=1/8, P(bar(E_(1))nnE_(2)nnbar(E_(3)))=1/8 . Find P(bar(E_(3)))

If E_1 and E_2 are independent events and P(E_1)=rho_1 , P(E_2)=rho_2 then find P(E_1^c cap E_2^c) .

If E_1 and E_2 are independent events such that P(E_1)=1/3 and P(E_2)=1/6 find P(E_1^c cap E_2^c) .