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9.4 · Q56

Q.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\frac{2}{5}, 12\frac{1}{2}, and 23\frac{2}{3} respectively. The probability of opening the messages by Group I, Group II and Group III are 12\frac{1}{2}, 14\frac{1}{4} and 14\frac{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.

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P(sports)=12(25)+14(12)+14(23)=15+18+16P(\text{sports})=\frac{1}{2}(\frac{2}{5})+\frac{1}{4}(\frac{1}{2})+\frac{1}{4}(\frac{2}{3})=\frac{1}{5}+\frac{1}{8}+\frac{1}{6}. With LCM 120: 24120+15120+20120=59120\frac{24}{120}+\frac{15}{120}+\frac{20}{120}=\frac{59}{120}. By Bayes' theorem, $P(\text{III}/\text{sports})=\dfrac{(1/4)(2/3)}{59/12 …

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