A dice is thrown three times. If the first throw shows 4, then the probability of getting 15 as the sum is:

$\frac{1}{18}$

Dice is thrown 3 times. First outcome is 4.

The cases for getting the sum as 15: (4,5,6) (4,6,5) (4,2,5) (4,5,2) (4,3,4) (4,4,3) ( = 6 Cases

Total number of outcomes with first number as 4 = 6×6 = 36

Required Probability = \(\frac{2}{36}\) = \(\frac{1}{18}\)

The correct answer is Option (3) → $\frac{1}{18}$