Two dice are thrown simultaneously. Find the probability of getting
(i) A total of at least 10
(ii) A doublet of even number
(iii) Odd number on both dice
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Step-by-step explanation:
Total number of possible cases =36
Favourable cases of getting a total of at least 10
={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}
Total number of favourable cases =6
P(Total of at least 10) =
36
6
=
6
1
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