Answer:
1)5/36
2)1/9
3)11/36
4)8/9
5)1/6
Explanation:
E: throwing two dice
Two dice thrown in a single toss,
in this event the outcomes may
be (1,1),(1,2).....(1,6),(2,1).....(2,6),(3,1).......(6,6)
total 6×6 outcomes: 36 outcomes
here,total possible outcome is 36
P(of anything)=(Total no. of favourable outcome/total possible outcome)
now, (1)sum of 8
we get sum of 8 only when it comes
(2,6),(3,5),(4,4),(5,3),(6,2)
Total no. of favourable outcome is 5
P(getting sum of 8)= 5/36
2) sum of 5
we get sum of 5 only when it comes
(1,4),(2,3),(3,2),(4,1)
Total no. of favourable outcome is 4
P(getting sum of 5)= 4/36=1/9
3)sum of 7 or 8
we get sum of 8 only when it comes
(2,6),(3,5),(4,4),(5,3),(6,2)
or, get 7 when it comes
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Total no. of favourable outcome is 11
P(getting sum of 7 or 8)= 11/36
4)sum to be not 9
we get sum of 9 only when it comes
(3,6),(4,5),(5,4),(6,3)
Total no. of favourable outcome is 4
P(getting sum of 9)= 4/36=1/9
here,we have to find probability of getting (not 9),
which is equal to : 1- probability of getting 9
=1 -1/9= 8/9
5) At least 10
atleeast 10 includes 10 to maximum (12)
i.e, 10,11&12
fav outcome: (4,6),(5,5)(6,4) & (5,6),(6,5) & (6,6)
total no. of fav outcome is 6
probability (atleast 10)=6/36 =1/6
✌️:)