- #1
davedave
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Here is the problem.
If you roll 5 six-sided dice at the same time, what is the probability that the sum of the roll is 20?
This is the solution that someone did.
The problem can be viewed as distributing 20 identical objects into 5 containers so that each has between 1 and 6 objects inclusive.
Each container must have at least 1 object because each die shows 1 as the smallest number. This reduces to distributing 15 objects.
There are C(15+5-1,15) ways to distribute 15 objects to the 5 containers.
The condition is that each die shows less than or equal to 6.
(1) There are C(5,1)=5 ways to satisfy the condition and there are C(9+5-1,9) combinations.
(2) There are C(5,2)=10 ways to satisfy the condition and there are C(3+5-1,3)
combinations.
So, [C(15+5-1,15)+10*C(3+5-1,3)-5*C(9+5-1,9)]/6^5 = 651/7776
This is the correct answer.
I am very confused about the statements (1) and (2) above.
(a) Why are there C(5,1) and C(5,2) ways to satisfy the condition?
(b) Why are there C(9+5-1,9) and C(3+5-1,3) combinations?
(c) Why do you have to subtract 5*C(9+5-1,9) in the final step of the solution?
Could someone please explain? I would really appreciate your help.
If you roll 5 six-sided dice at the same time, what is the probability that the sum of the roll is 20?
This is the solution that someone did.
The problem can be viewed as distributing 20 identical objects into 5 containers so that each has between 1 and 6 objects inclusive.
Each container must have at least 1 object because each die shows 1 as the smallest number. This reduces to distributing 15 objects.
There are C(15+5-1,15) ways to distribute 15 objects to the 5 containers.
The condition is that each die shows less than or equal to 6.
(1) There are C(5,1)=5 ways to satisfy the condition and there are C(9+5-1,9) combinations.
(2) There are C(5,2)=10 ways to satisfy the condition and there are C(3+5-1,3)
combinations.
So, [C(15+5-1,15)+10*C(3+5-1,3)-5*C(9+5-1,9)]/6^5 = 651/7776
This is the correct answer.
I am very confused about the statements (1) and (2) above.
(a) Why are there C(5,1) and C(5,2) ways to satisfy the condition?
(b) Why are there C(9+5-1,9) and C(3+5-1,3) combinations?
(c) Why do you have to subtract 5*C(9+5-1,9) in the final step of the solution?
Could someone please explain? I would really appreciate your help.
