Three critics review a book. For the three cr | vedclass

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Question

(C) Let $P(C_1), P(C_2), P(C_3)$ be the probabilities that the three critics are in favour of the book. Given the odds in favour are $(5: 2), (4: 3),

Vedclass · Accepted Answer

$\frac{209}{343}$

Vedclass · Answer

(C) Let $P(C_1), P(C_2), P(C_3)$ be the probabilities that the three critics are in favour of the book. Given the odds in favour are $(5: 2), (4: 3), (3: 4)$,we have: $P(C_1) = \frac{5}{5+2} = \frac{5}{7}, P(\bar{C}_1) = \frac{2}{7}$ $P(C_2) = \frac{4}{4+3} = \frac{4}{7}, P(\bar{C}_2) = \frac{3}{7}$ $P(C_3) = \frac{3}{3+4} = \frac{3}{7}, P(\bar{C}_3) = \frac{4}{7}$ For the majority to be in favour,at least two critics must be in favour. Required probability $= P(C_1)P(C_2)P(\bar{C}_3) + P(C_1)P(\bar{C}_2)P(C_3) + P(\bar{C}_1)P(C_2)P(C_3) + P(C_1)P(C_2)P(C_3)$ $= (\frac{5}{7} 	imes \frac{4}{7} 	imes \frac{4}{7}) + (\frac{5}{7} 	imes \frac{3}{7} 	imes \frac{3}{7}) + (\frac{2}{7} 	imes \frac{4}{7} 	imes \frac{3}{7}) + (\frac{5}{7} 	imes \frac{4}{7} 	imes \frac{3}{7})$ $= \frac{80}{343} + \frac{45}{343} + \frac{24}{343} + \frac{60}{343} = \frac{209}{343}$

Three critics review a book. For the three critics,the odds in favour of the book are $(5: 2)$,$(4: 3)$ and $(3: 4)$ respectively. The probability that the majority is in favour of the book is

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