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Suppose you toss a pair of dice 10 times. What is the probability of getting the following outcome: two rolls of 7, two rolls of 6, and any other outcome on the remaining six rolls?

yuhki  Nov 21, 2014

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 #1
avatar+17721 
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There are 6 · 6 outcomes for tossing two dice.

Of these ways there are 6 ways to get a sum of 7:  1&6, 2&5, 3&4, 4&3, 5&2, and 6&1.

There are 5 ways to get a sum of 6:  1&5, 2&4, 3&3, 4&2, and 5&1.

The probability of getting a 7 from a roll of two dice is 6/36.    (Let me call this outcome A.)

The probability of getting a 6 from a roll of two dice is 5/36.    (Let me call this outcome B.)

The probability of getting any other outcome is 25/36.            (Let me call this outcome X.)

The number of ways to arrange AABBXXXXXX is 10! / [ 2! · 2! · 6! ]  =  1260 ways 

So the probability is   1260 · (6/36)² · (5/36)² · (25/36)^6  =  0.0198 (approximately) 

geno3141  Nov 22, 2014
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1+0 Answers

 #1
avatar+17721 
+5
Best Answer

There are 6 · 6 outcomes for tossing two dice.

Of these ways there are 6 ways to get a sum of 7:  1&6, 2&5, 3&4, 4&3, 5&2, and 6&1.

There are 5 ways to get a sum of 6:  1&5, 2&4, 3&3, 4&2, and 5&1.

The probability of getting a 7 from a roll of two dice is 6/36.    (Let me call this outcome A.)

The probability of getting a 6 from a roll of two dice is 5/36.    (Let me call this outcome B.)

The probability of getting any other outcome is 25/36.            (Let me call this outcome X.)

The number of ways to arrange AABBXXXXXX is 10! / [ 2! · 2! · 6! ]  =  1260 ways 

So the probability is   1260 · (6/36)² · (5/36)² · (25/36)^6  =  0.0198 (approximately) 

geno3141  Nov 22, 2014

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