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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