Asked by keith
A die is thrown twice. Find the probability that the number obtained in the first throw is smaller than that in the second throw.
Answers
Answered by
Steve
just add them up, considering the 1st throw:
6: 0
5: 1/6
4: 2/6
...
1: 5/6
sum: (1/6)(15/6) = 5/12
since the chance of throwing each of the 1st numbers is 1/6.
6: 0
5: 1/6
4: 2/6
...
1: 5/6
sum: (1/6)(15/6) = 5/12
since the chance of throwing each of the 1st numbers is 1/6.
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