Suppose a six sided die has sides numbered 1-6. IF a person throws the die two times what is the probabiloty that the second number will be larger then the first?
Larger as in strictly larger (not equal to).
There are 36 possible different throws.
And we need to find all possible outcomes that make what you said true.
If he throws 1 first: there are 2,3,4,5,6=5 different outcomes.
If he throws 2 first: there are 3,4,5,6=4 different outcomes.
....
If he throws 5 first: there are 6=1 different outcomes.
If he throws 6 first: there are 0 different outcomes.
Thus,
5+4+3+2+1=5(6)/2=15
Thus, the probability 15/36=5/12