.Two dice are thrown .find the probability of getting (a).A sum of 10 (b). a multiple of 3 as the sum (c).an even number on the first dice and an odd number on the second dice.
Answers
Step-by-step explanation:
Total number of outcomes =6
2
=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
36
6
=
6
1
4) Favourable case of doublet of even nos.
=(2,2),(4,4),(6,6)
No. of cases=3
Required probability =
36
3
=
12
1
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 =
36
12
=
3
1