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Question: If \({{E}_{1}}\) and \({{E}_{2}}\) are equally likely, mutually exclusive and exhaustive events and ...

If E1{{E}_{1}} and E2{{E}_{2}} are equally likely, mutually exclusive and exhaustive events and P(A/E1)=0.2P\left( A/{{E}_{1}} \right)=0.2 and P(A/E2)=0.3P\left( A/{{E}_{2}} \right)=0.3 . Then find P(E1/A)P\left( {{E}_{1}}/A \right) .

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Solution

Here in this question we have been given that there are two events E1{{E}_{1}} and E2{{E}_{2}} are equally likely, mutually exclusive and exhaustive events and P(A/E1)=0.2P\left( A/{{E}_{1}} \right)=0.2 and P(A/E2)=0.3P\left( A/{{E}_{2}} \right)=0.3 . We have been asked to find P(E1/A)P\left( {{E}_{1}}/A \right) . We will use Bayes theorem which states that when we have E1,E2,.......,En{{E}_{1}},{{E}_{2}},.......,{{E}_{n}} a set of events associated with a sample space SS , where all the events E1,E2,.......,En{{E}_{1}},{{E}_{2}},.......,{{E}_{n}} have nonzero probability of occurrence and they form a partition of SS . Let AA be any event associated with SS , then according to Bayes theorem, P(Ei/A)=P(Ei)P(A/Ei)∑P(Ek)P(A/Ek)P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum{P\left( {{E}_{k}} \right)P\left( A/{{E}_{k}} \right)}} for any k=1,2,..........,nk=1,2,..........,n .

Complete step by step answer:
Now considering from the question we have been given that there are two events E1{{E}_{1}} and E2{{E}_{2}} are equally likely, mutually exclusive and exhaustive events and P(A/E1)=0.2P\left( A/{{E}_{1}} \right)=0.2 and P(A/E2)=0.3P\left( A/{{E}_{2}} \right)=0.3 .
We have been asked to find P(E1/A)P\left( {{E}_{1}}/A \right) .
From the basic concepts of probability we know that Bayes theorem is stated as when we have E1,E2,.......,En{{E}_{1}},{{E}_{2}},.......,{{E}_{n}} a set of events associated with a sample space SS , where all the events E1,E2,.......,En{{E}_{1}},{{E}_{2}},.......,{{E}_{n}} have nonzero probability of occurrence and they form a partition of SS . Let AA be any event associated with SS , then according to Bayes theorem, P(Ei/A)=P(Ei)P(A/Ei)∑P(Ek)P(A/Ek)P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum{P\left( {{E}_{k}} \right)P\left( A/{{E}_{k}} \right)}} for any k=1,2,..........,nk=1,2,..........,n .
Here we have been given that the events E1{{E}_{1}} and E2{{E}_{2}} are equally likely, mutually exclusive and exhaustive events. Therefore P(E1)=P(E2)=0.5P\left( {{E}_{1}} \right)=P\left( {{E}_{2}} \right)=0.5 .
Now by applying Bayes theorem we will have
P(E1/A)=P(E1)P(A/E1)P(E1)P(A/E1)+P(E2)P(A/E2) ⇒0.5(0.2)0.5(0.2)+0.5(0.3) ⇒25⇒0.4 \begin{aligned} & P\left( {{E}_{1}}/A \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( A/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A/{{E}_{2}} \right)} \\\ & \Rightarrow \dfrac{0.5\left( 0.2 \right)}{0.5\left( 0.2 \right)+0.5\left( 0.3 \right)} \\\ & \Rightarrow \dfrac{2}{5}\Rightarrow 0.4 \\\ \end{aligned}
Therefore we can conclude that P(E1/A)=0.4P\left( {{E}_{1}}/A \right)=0.4 .

Note: While answering questions of this type we should be sure with our concepts that we are going to apply and the calculations that we are going to perform in between the steps. Here generally we make mistakes during the calculation part so be careful.