For mutually exclusive events $R_{1}, R_{2}$, and $R_{3}$, we have $P\left(R_{1}\right)=0.05$, $P\left(R_{2}\right)=0.5$, and $P\left(R_{3}\right)=0.45$. Also, $P\left(Q \mid R_{1}\right)=0.7$, $P\left(Q \mid R_{2}\right)=0.6$, and $P\left(Q \mid R_{3}\right)=0.6$. Find $P\left(R_{1} \mid Q\right)$. \[ P\left(R_{1} \mid Q\right)= \] (Type an integer or a simplified fraction.)

Step 1 ：We are given the following probabilities: $P(R_{1}) = 0.05$, $P(R_{2}) = 0.5$, $P(R_{3}) = 0.45$, $P(Q|R_{1}) = 0.7$, $P(Q|R_{2}) = 0.6$, and $P(Q|R_{3}) = 0.6$.

Step 2 ：We want to find $P(R_{1}|Q)$, and we can use Bayes' theorem to do this. Bayes' theorem states that $P(A|B) = \frac{P(B|A) * P(A)}{P(B)}$.

Step 3 ：First, we need to find $P(Q)$, which is the probability of $Q$ occurring given all possible events. Since $R_{1}$, $R_{2}$, and $R_{3}$ are mutually exclusive and exhaustive, we have $P(Q) = P(Q|R_{1})P(R_{1}) + P(Q|R_{2})P(R_{2}) + P(Q|R_{3})P(R_{3})$.

Step 4 ：Substituting the given values, we find that $P(Q) = 0.7*0.05 + 0.6*0.5 + 0.6*0.45 = 0.605$.

Step 5 ：Now we can substitute $P(Q)$ into Bayes' theorem to find $P(R_{1}|Q)$. So, $P(R_{1}|Q) = \frac{P(Q|R_{1}) * P(R_{1})}{P(Q)}$.

Step 6 ：Substituting the given values, we find that $P(R_{1}|Q) = \frac{0.7*0.05}{0.605} = 0.05785123966942148$.

Step 7 ：Rounding this to three decimal places, we get $P(R_{1}|Q) = 0.058$.

Step 8 ：As a simplified fraction, this is $\frac{58}{1000}$, which simplifies further to $\frac{29}{500}$.

Step 9 ：So, the final answer is $P(R_{1}|Q) = \boxed{\frac{29}{500}}$.