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

Probability that the first critic favours the book $P\left(E_1\right)=\frac{5}{5+2}=\frac{5}{7}$

Probability that the second critic favours the book $P\left(E_2\right)=\frac{4}{4+3}=\frac{4}{7}$

Probability that the third critic favours the book $P\left(E_3\right)=\frac{3}{3+4}=\frac{3}{7}$

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

$\begin{array}{l}=P\left(E_1\cap E_2\cap {\overline{E}}_3\right)+P\left(E_1\cap {\overline{E}}_2\cap E_3\right)+P\left({\overline{E}}_1\cap E_2{E}_3\right)+P\left(E_1\cap E_2\cap E_3\right)\\ =P\left(E_1\right)P\left(E_2\right)P\left({\overline{E}}_3\right)+P\left(E_1\right)P\left({\overline{E}}_2\right)P\left(E_3\right)+P\left({\overline{E}}_1\right)P\left(E_2\right)P\left(E_3\right)+P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)\\ =\frac{5}{7}\times \frac{4}{7}\times \left(1-\frac{3}{7}\right)+\frac{5}{7}\times \left(1-\frac{4}{7}\right)\times \frac{3}{7}+\left(1-\frac{5}{7}\right)\times \frac{4}{7}\times \frac{3}{7}+\frac{5}{7}\times \frac{4}{7}\times \frac{3}{7}\\ =\frac{209}{343}\end{array}$