Events $A_{1}$ and $A_{2}$ are mutually exclusive and form a complete partition of a sample space $S$ with $P\left(A_{1}\right)=0.43$. If $E$ is an event in $S$ with $P\left(E \mid A_{1}\right)=0.27$ and $P\left(E \mid A_{2}\right)=0.1$, compute $P\left(A_{1} \mid E\right)=$ ? (Hint: Because $A_{1}$ and $A_{2}$ are mutually exclusive and form a complete partition of the sample space, $P\left(A_{2}\right)=1-P\left(A_{1}\right)$ ).

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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 \(\boxed{0.6707}\).