I have searched for similar answers with no luck so I am hoping one of you can be kind enough to help me with this.

Using standard 6 sided die - all results should be based on only one roll for each question of the following number of die. Each question below is independent of other questions, they are only based on one roll with n:

1 - Roll 5 die - probability of rolling three 1s with the other two die being any number

2 - Roll 4 die - probability of rolling three 1s with the other die being any number

3 - Roll 3 die - probability of rolling three 1s

4 - Roll 5 die - probability of rolling two 1s with the other three die being any number

5 - Roll 4 die - probability of rolling two 1s with the other two die being any number

6 - Roll 3 die - probability of rolling two 1s with the other die being any number

7 - Roll 2 die - probability of rolling two 1s

8 - Roll 5 die - probability of rolling five 1s

9 - Roll 4 die - probability of rolling four 1s


Along with the answers, if you could post the formula you used to calcuate this, I would appreciate it.

If there is a missing factor in the question, please answer it with what you used for that missing factor.

thanks !

Wow, that's not an easy one. Just keep checking back, I'm sure someone will come along that can help.

I have searched for similar answers with no luck so I am hoping one of you can be kind enough to help me with this.

Using standard 6 sided die - all results should be based on only one roll for each question of the following number of die. Each question below is independent of other questions, they are only based on one roll with n:

These appear to be binomial in nature.

The general form is C(n,k)p^{k}q^{n-k}, where q=1-p.

Normally, there would be a sum sign in front, but you do not need that since they are not asking for 'at least' or 'at most' or something like that.


1 - Roll 5 die - probability of rolling three 1s with the other two die being any number

C(5,3)(\frac{1}{6})^{3}(\frac{5}{6})^{2}=\frac{125 }{3888}\approx .032

You could also think of it as prob. Of rolling 3 ones is (\frac{1}{6})^{2}

The prob. Of rolling the 2 anything else is (\frac{5}{6})^{2}

So, multiply them (\frac{1}{6})^{3}(\frac{5}{6})^{2}

But these can be arranged in \frac{5!}{3!2!}=10 ways.

So, we have (\frac{1}{6})^{3}(\frac{5}{6})^{2}(10)=\frac{125}{ 3888}

Same as before.

Try it with the others. They're very similar


Probability of rolling 5 ones with 5 die

C(5,5)(\frac{1}{6})^{5}(\frac{5}{6})^{0}