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

A
35/49
B
125/343
C
164/343
D
209/343

Solution:

The probability that the first critic favorer the book is $P (E_{1}) = \frac{5}{5 + 2} = \frac{5}{7}$

The probability that the second critic favors the book is $P (E_{2}) = \frac{4}{4 + 3} = \frac{4}{7}$
The probability that the third critic favors the book is $P (E_{3}) = \frac{3}{3 + 4} = \frac{3}{7}$
Majority will be in favor of the book, if atleast two critics favor the book.
Hence, the probability is

$P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3})$

$= P (E_{1}) P (E_{2}) P ({\bar{E}}_{3}) + P (E_{1}) P ({\bar{E}}_{2}) P (E_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1}) P (E_{2}) P (E_{3})$

$= \frac{5}{7} \times \frac{4}{7} \times (1 - \frac{3}{7}) + \frac{5}{7} \times (1 - \frac{4}{7}) \times \frac{3}{7} + (1 - \frac{5}{7}) \times \frac{4}{7} \times \frac{3}{7} + \frac{5}{7} \times \frac{4}{7} \times \frac{3}{7} = \frac{209}{343}$

Solution:

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