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

The probability that the first critic favours the book is

$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{5}{5+2}=\frac{5}{7}$

The probability that the second critic favours the book is

$\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{4}{4+3}=\frac{4}{7}$

The probability that the third critic favours 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 at least two critics favor the book. Hence, the probability is

$P\left(E_1\cap E_2\cap {\overline{E}}_3\right)+P\left(E_1\cap {\overline{E}}_2\cap E_3\right)+P\left({\overline{E}}_1\cap E_2\cap E_3\right)+P\left(E_1\cap E_2\cap E_3\right)$

$=P\left(E_1\right)P\left(E_2\right)P\left({\overline{E}}_3\right)+P\left(E_1\right)P\left({\overline{E}}_2\right)P\left(E_3\right)+P\left({\overline{E}}_1\cap E_2\cap E_3\right)+P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)$

$$\begin{gathered} =\frac{5}{7}\times \frac{4}{7}\times \left(1-\frac{3}{7}\right)+\frac{5}{7}\times \left(1-\frac{4}{7}\right)\times \frac{3}{7}+\left(1-\frac{5}{7}\right)\times \frac{4}{7}\times \frac{3}{7}+\frac{5}{7}\times \frac{4}{7}\times \frac{3}{7} \\ =\frac{209}{343} \end{gathered}$$