Question

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Answer 1

2 Dices are thrown
The outcomes are :-

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Hence the total number of outcomes are 6×6
= 36

Number of outcomes which have even sum :-

 (1 , 1 ) (1 , 3 ) ( 1 , 5 )
( 2 , 2 ) ( 2 , 4 ) ( 2 , 6 )
 ( 3 , 1 ) ( 3 , 3 ) ( 3 , 5 )
( 4 , 2 ) ( 4 , 4 ) ( 4 , 6)
( 5 , 1 ) ( 5 , 3 ) ( 5 , 5 )
 ( 6 , 2 ) ( 6 , 4 ) ( 6 , 6 )

P(E) = Number of favourable outcomes
                    Total no. of possible outcomes
               =  18 / 36
               =  1 / 2.

Number of outcomes which have even product :-

    ( 1 , 2 ) ( 1, 4 ) ( 1 , 6 )
    ( 2 , 1 )  ( 2 , 2 ) ( 2 , 3 ) ( 2 , 4 ) ( 2 , 5 ) ( 2 , 6 )
    ( 3 , 2 ) ( 3 , 4 ) ( 3 , 6 )
    ( 4 , 1 ) ( 4 , 2 ) ( 4 , 3 ) ( 4 , 4 ) ( 4 , 5 ) ( 4 , 6 )
    ( 5 , 2 ) ( 5 , 4 ) ( 5 , 6 )
    ( 6 , 1 ) ( 6 , 2 ) ( 6 , 3 ) ( 6 , 4 ) ( 6 , 5 ) ( 6 , 6 )

     P(E)  =   Number of favorable outcomes  
                         Total number of possible outcomes
                      =  27 / 36
                      = 3 / 4.

Hope this helps!!
cheers!! (:

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