How many ways to get exactly one 6 on 4 rolls of a dice (without using negation) I've started to learn probability, and the first thing I saw was the question about the probability of getting at least one six in $4$ rolls of a dice.
I understand that it's easier to do $1-(\frac{5}{6})^4$ because it says "at least", but what if it said the probability of getting exactly one six in those $4$ rolls?
Would it just be ${4\choose 1}(\frac{1}{6})(\frac{5}{6})^3$?
Because you have to choose which of the four rolls will have the 6, and the one that has the six as a $\frac{1}{6}$ chance and the other 3 have a $\frac{5}{6}$ chance?
 A: Your title says exactly one six and your question says at least one.  Your calculation is correct for exactly one.  $1-(\frac 56)^4$ computes at least one.  Which are you trying to compute?
A: It sounds like you are asking two questions.


*

*Let $A = \{\text{At least one 6}\}$. Then (with no negation)
$$P(A) = \sum_{k=1}^4 \binom{4}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4-k}.$$

*Let $B =\{\text{Exactly one 6}\}$. Then
$$P(B) = \binom{4}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^3.$$
A: The probability of getting a specific value in exactly $m$ out of $n$ rolls:
$$\binom{n}{m}\cdot\left(\frac16\right)^{m}\cdot\left(\frac56\right)^{n-m}$$

The probability of getting a specific value in at least $m$ out of $n$ rolls:
$$\sum\limits_{k=m}^{n}\binom{n}{k}\cdot\left(\frac16\right)^{k}\cdot\left(\frac56\right)^{n-k}$$

The probability of getting a specific value in at most $m$ out of $n$ rolls:
$$\sum\limits_{k=0}^{m}\binom{n}{k}\cdot\left(\frac16\right)^{k}\cdot\left(\frac56\right)^{n-k}$$
