Question

A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15

Answer 1

HEYAA..!!
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There are 3 positions to be filled.
____+ ____ +____ = 15

The possible combinations are 
Case 1: (6 6 3)
(6 3 6)
(3 6 6) total 3 combinations

Case 2: (4 5 6)
(4 6 5)
(5 6 4)
(5 4 6)
(6 5 4)
(6 4 5) total 6 combinations

Case 3: (5 5 5) only 1 combination

So probability is possible/total

Possible is 2 because, the question says that uppermost face should show 4, and in case 2 only the 4 appears twice as the uppermost face so it will be 2

Total possibilities would be 6 + 3+ 1 = 10

So probability is 2/10 or 1/5
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:-)

Answer 2

The sum of numbers can be 15 in the following three ways :

Case I: 15=3+6+6
The first, second and third throws can be (3,6,6),(6,3,6)and (6,6,3) respectively.
Total number of ways in which 3,6 and 6 can be obtained = 6

Case II: 15=4+5+6
The first, second and third throws can be  4, 5 and 6.
Total number of ways in which 4,5 and 6 can be obtained = 6

Case III: 15=5+5+5
The first, second and third throws can be 5,5 and 5.
Total number of ways in which 5,5, and 5 can be obtained = 1.

Hence, The total number of ways=3+6+1=10
The total number of ways in which the first roll will be 4 is 2.

Required chance = 2/10 = 1/5