# 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 25,12, and 23 respectively. - Mathematics and Statistics

Sum

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 E1, E2, E3 be the events that Group I, Group II, and Group III open the messages

Then, P(E1) = 1/2, P(E2) = 1/4, P(E3) = 1/4

E1, E2, E3 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

Concept: Baye'S Theorem
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