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A pair of dice is thrown simultaneously.Find   (i) P(getting a prime no. each) (ii)P(not getting prime no. on each die)

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1 Prime numbers 2 , 3 and 5 .Favourable event : 2,2 2,3 2,5 3,2 3,3 3,5 5,2 5,3 5,5 PGetting prime number each=936=142 PNot getting prime number each=1-PGetting prime number each=1-14=34 .
 
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hi,
now 
possible outcomes are  =36
        1      2      3     4         5      6
1  [1,1] [1,2] [1,3] [1,4] [1,5] [1,6]
2  [2,1] [2,2] [2,3] [2,4] [2,5] [2,6]
3  [3,1] [3,2] [3,3] [3,4] [3,5] [3,6]
4  [4,1] [4,2] [4,3] [4,4] [4,5] [4,6]
5  [5,1] [5,2] [5,3] [5,4] [5,5] [5,6]
6  [6,1] [6,2] [6,3] [6,4] [6,5] [6,6]
total outcomes = 36
1] P[ getting prime no . on each } = 8/36 = 2/9 [ 2,3 & 5,3  prime no 2,3 & 5 ]
2] P [not getting on each die ] = 1 - 2/9 = 9-2/9 = 7/9 
HOPE IT MAY HELP U  AND BEST OF LUCK FOR UR EXAM
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