SOLUTION: Suppose that two dice are tossed. For each die, it is equally likely that 1, 2, 3, 4, 5, or 6 dots will turn up. Let S be the sum of the two dice. a. Let E1 be the event that the

(a)  We should check if  P (E1 & E2) = P(E1)*P(E2).


     Calculation P(E1) is easy:  P(E1) = ,  since there are 6 outcomes for the first die, 
     showing the numbers 1, 2, 3, 4, 5 and 6 with equal probabilities.


     To calculate P(E2), notice that the sum of 6 of two dice can be obtained in 5 (five)  cases

         (1,5), (2,4), (3,3), (4,2) and (5,1),

     So the event E2 has the probability  P(E2) = .



     Now, the event (E1 & E2)  is the intersection of E1 and E2; 

     in other words, first die shows a "3", while the sum of the faces of the two dice is  6.


     It may happen if and only if BOTH dice face up a "3" -- the event with the probability  ,  obviously.


     Thus we have  P(E1) = ;  P(E2) =   and  P(E1 & E2) = .


     The product of probabilities P(E1)*P(E2) = } =  is not equal to  P(E1 & E2) =  --

     hence, the two events E1 and E2 ARE NOT INDEPENDENT.    ANSWER




(b)  If the first die faces up a "3", then the sum is even in these  3 (three)  cases

         (3,1), (3,3), (3,5).

     So the probability for S to be an even number is   = .    ANSWER