Throwing a die 7 times

[binjip]

Hello,

my problem is as follows:

Throw a die 7 times.

i) What is the probability that you get number 6 twice and all other numbers (1,2,3,4,5) once. (e.g. one possible set of outcomes would be 6, 6, 5, 4, 3, 2, 1)

ii) What is the probability that you get all the numbers of a die? (e.g. 6, 5, 4, 3, 2, 1, x where x is {1, ..., 6})

Attempt at solution:

i)

There are 6^7 possible ways of arranging the set of 7 outcomes (# of permutations).

There 6*5*4*3*2*1*1 = 6! ways of arranging the numbers under given conditions.

P(two sixes, and all other outcomes) = 6! / 6^7 = 5! / 6^6

ii) # of permutations is 6^7. No change here.

The # of ways to arrange the numbers under given conditions changes to:

6*6*5*4*3*2*1 = 6*6!

P(all 6 numbers of a die) = 6*6! / 6^7 = 5! / 6^5


Is this correct? Would appreciate any comments. Thanks

 

[Dason]

For question i - Why do you think the number of ways to get two sixes and the other 5 as anything else is (6!)?

 

[binjip]

Thanks Dason for your answer.

Well, suppose we had only 6 throws. The possible ways to have the first number is 6 (there are 6 sides to a die). There are 5 ways to get the 2nd, 3 the 3rd, ..., 1 the last 6th one. This is 6!.

Now I argued (and I think now this was a mistake), that I have an additional 7th throw in which I needed a 6 (so I multiplied with 1).

However, I overlooked that the 6 can be thrown in any of the 7 turns, so I would multiply with 7 instead (there are 7 additional possible ways to arrange the numbers).

This would change the result to 7! / 6^7

Would you agree?

 

[BGM]

Multinomial Distribution.

 

[binjip]

Oh right, I see it now.

The problem with throwing a 6 twice and 1,...,5 once each is C(7;1!,1!,1!,1!,1!,2!). Thanks.