Question

if a dice is marked with 1 ,2,3,4,5 and 6 on its faces what is the probability of getting a three when thrown ​

Answer 1

Answer:

There are 3 ways that the end two dice total 4

1 on the first die + 3 on the seventh die = 4

2 on the first die + 2 on the seventh die = 4

3 on the first die + 1 on the seventh die = 4

Therefore the First die can be any of 3

The 7th die has no choice. If the first die is a 1 it can only be a 3, if the first die is a 2, the 7th can only be a 2, if the first die is a 3, it can only be a 1. Therefore there is no choice for the 7th die, it has to be whatever makes the sum 4 given the first die's value

The second, third, 4th, 5th and 6th dies though can be any of the 6 numbers 1 - 6. Therefore there are 6 choices for all of these

Total number of arrangements is therefore

3 * 6 * 6 * 6 * 6 * 6 * 1 = 23328 different arrangements

Answer = 23328