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A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. The chance that the first roll was a four is
There are three possibilities for sum of uppermost faces of dice is equal to 15. They are
5, 5, 5 - only 1 arrangement
4, 5, 6 - 3! = 6 arrangements
3, 6, 6 - $$\frac{3!}{2}$$ = 3 arrangements
Out of these 10 arrangements, there are 2 cases in which 4 occurs in first roll ( 4,5,6 or 4,6,5)
Therefore, probability = $$\frac{2}{10}$$ = $$\frac{1}{5}$$
Answer is option B.
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