Thread: Probability--Rolling a die

This probably should be very easy but i am not good at probability.

Given a fair die, what is the probability of rolling at least 2 sixes if the die is rolled 12 times.

I have tried a little bit, and this is what I have.

First of all, i know the universe is equal to 6^12. And the hint is that it is easier to compute the probability of the event not happening.
If A is the event of getting at least 2 sixes, B is getting 1 or no sixes.

P(B)=?? Therefore, P(A)=1-P(B)


Thanks for any help.

Hello, dude15129!

You say you're not good at probability,
. . but you have a good understanding of it.

Your game plan is absolutely correct!


Number of ways to get no 6's: .

Number of ways to get one 6: .


Hence: .


Therefore: .

Thanks for your help,
but can you explain the way you got the number of ways to get 1 six? i don't understand the formula/(12 1) thing.


Also, the next part asks about the probability of rolling at least 3 sixes if the die is rolled 18 times. So, following what you did.... it would be....


P(rolling 0,1,2 sixes)= [(5/6)^18 + 18(1/6)(5/6)^17 + 18(1/6)(5/6)^16]/6^18


Thanks.

Originally Posted by dude15129

Also, the next part asks about the probability of rolling at least 3 sixes if the die is rolled 18 times. So, following what you did.... it would be....
[FONT=monospace]P(rolling 0,1,2 sixes)= [(5/6)^18 + 18(1/6)(5/6)^17 + 18(1/6)(5/6)^16]/6^

Are you sure that this is not fully explained in your text material?

If we roll a die 18 times, the the probability of getting exactly is given using the binominal formula.


That must be somewhere in your notes. If not, then you are not ready for this problem.

haha....well actually we haven't had any notes really and our textbook isn't like a real textbook.

but thanks for the help.