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Question

Two dice are thrown simultaneously. Find the probability of getting
1) an even number as the sum
2) the sum sum as a prime number
3) a total of at least 10
4) a doublet of even number
5) same number on both side
6) a multiple of 3 as the sum

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Solution

Total number of outcomes =62=36
1) Cases favourable to an even nos are the 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)=18cases
2) Cases favourable sum as prime nos
3(1,1),(1,2),(1,4),(1,6),(2,1)(2,3),(2,5),(3,2),(4,1),(5,2),(5,6),(6,1)(6,5)
No of cases =13
Required probability =13
3) Cases favourable to total of at least 10
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)
No. of cases =6
Required of probability 636=16
4) Favourable case of doublet of even nos.
=(2,2),(4,4),(6,6)
No. of cases=3
Required probability =336=112
5) Favourable cases as a multiple of 3 as the sum =(1,2),(1,5),(2,4),(2,6)=(3,3),(3,6),(4,2),(4,8)=(5,1),(5,4),(6,3),(6,6)
No. of cases =12
Required probability =1236=13

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