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least 6 from a dice

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by ruplun » Fri Jul 29, 2011 5:29 am
2 dices are thrown together . What is the probability of getting a total of at least 6?
a. 1/3
b. 1/2
c. 2/3
d. 13/18
e. 8/9
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by ColumbiaVC » Fri Jul 29, 2011 1:34 pm
D

Total prob=6x6=36
Total equal to 6 pairs are {(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
= 26

So, probability = 26/36=13/18

I am sure there would be a smarter way to deal with this problem. Looking forward for more explanations.
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by tryingtobeat » Fri Jul 29, 2011 2:58 pm
ColumbiaVC wrote:D

Total prob=6x6=36
Total equal to 6 pairs are {(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
= 26

So, probability = 26/36=13/18

I am sure there would be a smarter way to deal with this problem. Looking forward for more explanations.
I just did the opposite of what you did, I just found pairs that do not equal to at least six, which is 10. (36-10)/36 = 13/18
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by XLogic » Fri Jul 29, 2011 3:51 pm
ColumbiaVC wrote: I just did the opposite of what you did, I just found pairs that do not equal to at least six, which is 10...
Your method is faster. I tried this method at first, but lost my way... :)
All (Total < 6) = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
ColumbiaVC wrote: (36-10)/36 = 13/18
PS. I wonder if this question can be solved even faster... any takers?
my post helped --> thank me!
don't thank me --> my post = what the..??
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