There are three social media groups on a mobile: Group I, Group II and Group III. The probabilities that Group I, Group II and Group III sending the messages on sports are `2/5, 1/2`, and `2/3` respectively. The probability of opening the messages by Group I, Group II and Group III are `1/2, 1/4` and `1/4` respectively. Randomly one of the messages is opened and found a message on sports. What is the probability that the message was from Group III

#### Solution

Let E_{1}, E_{2}, E_{3} be the events that Group I, Group II, and Group III open the messages

Then, P(E_{1}) = `1/2`, P(E_{2}) = `1/4`, P(E_{3}) = `1/4`

E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive

Let M ≡ the event that message on sports is sent

`"P"("M"/"E"_1)` = Probability that sports message is sent given that it is from Group I

= `2/5`

Similarly, `"P"("M"/"E"_2) = 1/2, "P"("M"/"E"_3) = 2/3`

By Baye's Theorem, the required probability

= `"P"("E"_3/"M")`

`=("P"("E"_3)*"P"("M"/"E"_3))/("P"("E"_1)*"P"("M"/"E"_1) + "P"("E"_2)*"P"("M"/"E"_2) + "P"("E"_3)*"p"("M"/"E"_3))`

= `((1/4)*(2/3))/((1/2)*(2/5) + (1/4)*(1/2) + (1/4)*(2/3))`

= `((1/6))/((1/5) + (1/8) + (1/6))`

= `((1/6))/((59/120))`

= `20/59`