Three critics review a book. Odds in favour of the book are 5:2, 4:3, respectively, for the three critics. The probability that majority are in favour of the book is

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$\frac{125}{343}$

$\frac{35}{49}$

$\frac{209}{343}$

$\frac{164}{343}$

answer is D.

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Detailed Solution

Probability that the first critic favours the book $P (E_{1}) = \frac{5}{5 + 2} = \frac{5}{7}$

Probability that the second critic favours the book $P (E_{2}) = \frac{4}{4 + 3} = \frac{4}{7}$

Probability that the third critic favours the book $P (E_{3}) = \frac{3}{3 + 4} = \frac{3}{7}$

Majority will be in favour if at least two critics favour the book, i.e.,

$\begin{array}{l} = P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) + P ({\bar{E}}_{1} \cap E_{2} E_{3}) + P (E_{1} \cap E_{2} \cap E_{3}) \\ = P (E_{1}) P (E_{2}) P ({\bar{E}}_{3}) + P (E_{1}) P ({\bar{E}}_{2}) P (E_{3}) + P ({\bar{E}}_{1}) P (E_{2}) P (E_{3}) + P (E_{1}) P (E_{2}) P (E_{3}) \\ = \frac{5}{7} \times \frac{4}{7} \times (1 - \frac{3}{7}) + \frac{5}{7} \times (1 - \frac{4}{7}) \times \frac{3}{7} + (1 - \frac{5}{7}) \times \frac{4}{7} \times \frac{3}{7} + \frac{5}{7} \times \frac{4}{7} \times \frac{3}{7} \\ = \frac{209}{343} \end{array}$