What is the probability of getting
-atleast 1 six when 6 dice are rolled?
- atleast 2 six when 12 dice are rolled? - atleast 3 six when 18 dice are rolled?
"At least 1 six" is $1-(5/6)^6$. How can I do the next 2?
Can you please help me with some hints? I just don't get this.
On
The chance of no sixes in 12 dice, extending your result is $\left( \frac 56 \right)^{12}$. The chance of exactly one six is $12 \cdot \frac 16 \cdot \left( \frac 56 \right)^{11}$, where the $12$ is the number of ways to choose which die will be the $6$, the $\frac 16$ is the chance it is a $6$, the the $\frac 56$ is the chance that all the rest are non-$6$'s
At least two 6 when 12 dice are rolled is
1 - (probability that no 6 was rolled + probability that one 6 was rolled).
Using this, can you do the last one?
Edit: The probability of no 6 is $(5/6)^{12}$ (there are 12 dice). The probability of (exactly) one 6 is $12\times(1/6)\times(5/6)^{11}$: you have one die which is fixed at 6, all the other ones can be anything but 6, and you have 12 choices for which die has 6.