I encountered this question while solving few practice problems on probability theory. I have solved it in my own way, but am unsure about whether it's the correct way to go about it.
Four fair dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
My approach: $$ \begin{align*} P\big(\text{only 1 six}\big|\text{at least 1 six}\big) &= \frac{P(\text{only 1 six}, \text{ at least one six})}{P(\text{at least 1 six})} \\ &= \frac{P(\text{only 1 six})}{P(\text{at least 1 six}) } \end{align*} $$
Now, $P(\text{at least 1 six}) = 1 - \frac{5^3}{6^3}$ and $P(\text{only 1 six}) = ({}_{4}C_1) \frac{1}{6}$ $\frac{5^3}{6^3}$.
So, substituting, I get $\frac{500}{671}$ as my answer.
Is my approach correct? Any insight would be helpful.
$\textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$\textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let $B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $B\cap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique, $$ P(A)=1-P(A') = \boxed{1-(5/6)^4}. $$ Now, to find $P(B)$, we'll use the binomial probability distribution: $$ b(x;n,p) = \begin{cases} \binom{n}{x} p^x (1-p)^{n-x} &\mbox{ if }x = 0,1,2,\ldots, n, \\ \qquad\quad 0 & \mbox{ otherwise}, \end{cases} $$ where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$. So $$ P(B) = b(1; 4,1/6)= \binom{4}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{3} =\boxed{\frac{125}{324}}. $$ Putting this together, we have $$ P(B|A) = \frac{P(B\cap A)}{P(A)} = \frac{P(B)}{P(A)} = \color{green}{\boxed{\frac{500}{671}}}. $$