Choosing numbers at random with replacement

[AST]

Hello all,

I am trying to solve this problem:

Five numbers are being chosen from the set {1,2,3,4,5,6,7,8,9,10,11}, with replacement. Each number can be chosen 0 to 5 times.
What is the probability of getting 3 odd numbers ?

Solution:

It is easy to show that there are 11^5 options in this trial. The number of desired options is tricky. I initially thought that it's 6^3 * 5^2, but I think I am missing something. The odd numbers can be 111, or 113, or 311. I think that order matters, does it ?

Thank you for any assistance.

 

[blamocur]

I think that order matters, does it ?

It looked good to me at the first glance, but running a quick simulation showed the result which is 10 times larger -- want to guess why ? The hint is in your own question which I quoted.

 

[AST]

Is it (5 3) ?
Can you explain why ?

 

[Dr.Peterson]

Hello all,

I am trying to solve this problem:

Five numbers are being chosen from the set {1,2,3,4,5,6,7,8,9,10,11}, with replacement. Each number can be chosen 0 to 5 times.
What is the probability of getting 3 odd numbers ?

Solution:

It is easy to show that there are 11^5 options in this trial. The number of desired options is tricky. I initially thought that it's 6^3 * 5^2, but I think I am missing something. The odd numbers can be 111, or 113, or 311. I think that order matters, does it ?

Thank you for any assistance.

Is it (5 3) ?
Can you explain why ?

Yes; but can you explain why?

You have to choose which 3 of the 5 positions will hold an odd number. Then you can put odd numbers there, and even in the other two.

So, yes, it's C(5, 3)*6^3*5^2 for the numerator.