Step 1 :Given that events $A_{1}$ and $A_{2}$ are mutually exclusive and form a complete partition of a sample space $S$ with $P\left(A_{1}\right)=0.43$.
Step 2 :Also given that $E$ is an event in $S$ with $P\left(E \mid A_{1}\right)=0.27$ and $P\left(E \mid A_{2}\right)=0.1$.
Step 3 :Since $A_{1}$ and $A_{2}$ are mutually exclusive and form a complete partition of the sample space, we have $P\left(A_{2}\right)=1-P\left(A_{1}\right)=1-0.43=0.57$.
Step 4 :We can find $P(E)$ by using the law of total probability, which states that $P(E) = P(E | A_{1}) * P(A_{1}) + P(E | A_{2}) * P(A_{2})$. Substituting the given values, we get $P(E) = 0.27 * 0.43 + 0.1 * 0.57 = 0.1731$.
Step 5 :We are asked to find the conditional probability of event $A_{1}$ given event $E$. We can use Bayes' theorem to solve this problem. Bayes' theorem states that $P(A_{1} | E) = \frac{P(E | A_{1}) * P(A_{1})}{P(E)}$.
Step 6 :Substituting the known values into Bayes' theorem, we get $P(A_{1} | E) = \frac{0.27 * 0.43}{0.1731} \approx 0.6707$.
Step 7 :Final Answer: The probability of event $A_{1}$ given event $E$ is approximately \(\boxed{0.6707}\).