Show that the probability of not getting 7 or 11 total on either of two tosses of a pair of fair dice is 1/3.
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Show that the probability of not getting 7 or 11 total on either of two tosses of a pair of fair dice is 1/3.

[From: ] [author: ] [Date: 13-03-13] [Hit: ]
(4,3) , (3,4) , (2,5) ,......
I m not sure. Plz help

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Are you sure the answer is 1/3 ???

my answer is 7/9.

in the sample space of throwing 2 dice experiment[where total sample points are 36] the sample points to get 7 or 11 total are

(4,3) , (3,4) , (2,5) , (5,2), (6,1), (1,6), (6,5), (5,6) --hence 8 sample points

the probability of getting 7 or 11 total on either of two tosses of a pair of fair dice
=8/36 = 2/9

the probability of not getting 7 or 11 total on either of two tosses of a pair of fair dice
=1 - (2/9) = 7/9

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Well, if you just draw a 2 way table, you can see that the probability of getting a total of 7 or 11 total from 1 toss of a pair of dice is 8/36 = 2/9
So you toss them twice, with both of them not getting a total of 7 or 11, so (7/9)^2 = 49/81
I'm fairly sure whoever asked you this is incorrect.
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