Question

Two dice are thrown. Find the probability of getting the number on the upper face of the first die is less than the number on the upper face of the second die.

Answer 1

Favorable cases

(1,2) (1,3) (2,3) (1,4) (2,4) (3,4) (1,5) (2,5) (3,5)

(4,5) (1,6) (2,6) (3,6) (4,6) (5,6) = 15

Sample space $=36$

Probability of the number of first die to be less then second

$=\frac{Favorablecases}{Samplespace}$ $=\frac{15}{36}=\frac{5}{12}$