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

# 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

1. A

35/49

2. B

125/343

3. C

164/343

4. D

209/343

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### Solution:

The probability that the first critic favorer the book is $\mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{5}{5+2}=\frac{5}{7}$

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

$\mathrm{P}\left({\mathrm{E}}_{1}\cap {\mathrm{E}}_{2}\cap {\overline{\mathrm{E}}}_{3}\right)+\mathrm{P}\left({\mathrm{E}}_{1}\cap {\overline{\mathrm{E}}}_{2}\cap {\mathrm{E}}_{3}\right)+\mathrm{P}\left({\overline{\mathrm{E}}}_{1}\cap {\mathrm{E}}_{2}\cap {\mathrm{E}}_{3}\right)+\mathrm{P}\left({\mathrm{E}}_{1}\cap {\mathrm{E}}_{2}\cap {\mathrm{E}}_{3}\right)$

$=\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left({\overline{\mathrm{E}}}_{3}\right)+\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left({\overline{\mathrm{E}}}_{2}\right)\mathrm{P}\left({\mathrm{E}}_{3}\right)+\mathrm{P}\left({\overline{\mathrm{E}}}_{1}\cap {\mathrm{E}}_{2}\cap {\mathrm{E}}_{3}\right)+\mathrm{P}\left({\mathrm{E}}_{1}\right)\mathrm{P}\left({\mathrm{E}}_{2}\right)\mathrm{P}\left({\mathrm{E}}_{3}\right)$

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

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