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

#1**+5 **

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

#1**+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