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: .
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.
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.