As 3 dice are thrown. So total possible outcomes = 6×6×6 = 216
For getting 15 as the sum we need this combination-
(5, 5, 5) or (6,4,5) or (6,6,3)
(6, 4, 5) can be arranged in 3! = 6 ways.
(6, 4, 5) can be arranged in (3!)/(2!) = 3 ways
∴ 10 favourable outcomes are possible.
∴ P(E) = \(\frac{10}{216}\)
