EINSTIEN OBSERVATION ON P.E.E

A sample space consists of 9 elementary event E_1, E_2, E_3 ..... E_8, E_9 whose probabilities are P(E_1) = P(E_2) = 0. 08 ,P(E_3) = P(E_4) = 0. 1, P(E_6) = P(E_7) = 0. 2 ,P(E_8) = P(E_9) = 0. 07. Suppose A = {E_1,E_5,E_8}, B = {E_2, E_5, E_8, E_9}. Compute P(A), P(B) and P(AnnB). Using the addition law of probability, find P(AuuB). List the composition of the event AuuB, and calculate, P(AuuB) by adding the probabilities of the elementary events. Calculate P(barB) from P(B), also calculate P(barB) directly from the elementary events of barB.

Two events E and F are independent. If P(E )=0.3 and P(EcupF)=0.5 then P(E//F)-P(F//E) equals to

Let AansB be any two events with positive probabilities Statement I P((E)/(A))>=P((A)/(E))P(E) Statement I P((E)/(A))>=P(A nn E)

For any event E ,which of the following is correct P(E)>=0 P(E) 0 P(E)<0

= E_ (1), E_ (2), E_ (3), E_ (1), E_ (5) and P (E_ (1)) = (95) / (100), P (E_ (2) | E_ (1)) = (93) / (99), P (E_ (3) | E_ (1) E_ (2) E_ (3)) = (92) / (97) and P (E_ (5) | E_ (1) E_ (2) E_ (3) E_ (4)) = (91) / (96) n P (E) equals

If P(E) = 0.34, find p(not E).

If P(E)=0.37 then P (Not E) will be

Given that E and F are events such that P(E)=0.6,P(F)=0.3 and P(E nn F)=0.2 , find P(E/F) and P(F/E).