Question

1. Three dice are rolled. What is the probability that number on first die is less than the positive difference of second and third die?

Answer 1

When three dice are rolled then total 6*6*6=216 possible outcomes like as (1,1,1), (1,5,6), extra...

Given that the number on the first die is less than the positive difference of second and third die i.e.

favourable events are (1,1,3), (1,1,4), (1,1,5),(1,1,6), (1,2,4), (1,2,5), (1,2,6), (1,3,5), (1,3,6), (1,4,6), (2,1,4), (2,1,5), (2,1,6), (2,2,5),(2,2,6), (2,3,6), (3,1,5), (3,1,6), (3,2,6) ,(4,1,6).

Here one thing should be noted that condition of positive difference. There are two possibilities. One is 2nd number always greater than 3rd such that the difference between 2nd & 3rd is +ve or 3rd number always greater than 2nd such that the difference between 3rd & 2nd is +ve. I have condition 2 i.e. I supposed that 3rd number is always greater than the 2nd number.

Total no. of favourable cases= 20

Hence, the required probability= 20/216= 0.092