Question

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

• A. $\frac{35}{49}$
• B. $\frac{125}{343}$
• C. $\frac{164}{343}$
• D. $\frac{209}{343}$

Answer 1

The correct option is D $\frac{209}{343}$

The Probability that the first critic favors the book is
$P\left(E_1\right)=\frac{5}{5+2}=\frac{5}{7}$
The Probability that the second critic favors the book is
$P\left(E_2\right)=\frac{4}{4+3}=\frac{4}{7}$
The Probability that the third critic favors the book is
$P\left(E_1\right)=\frac{3}{3+4}=\frac{3}{7}$
Majority will be in favor of the book if at least two critics favor the book. Hence, the probability is
$P(E_1\cap E_2\cap {\overline{E}}_3)+P(E_1\cap {\overline{E}}_2\cap E_3)+P({\overline{E}}_1\cap E_2\cap E_3)+P(E_1\cap E_2\cap E_3)$
$=P\left(E_1\right)P\left(E_2\right)P\left({\overline{E}}_3\right)+P\left(E_1\right)P\left({\overline{E}}_2\right)P\left(E_3\right)+P\left({\overline{E}}_1\right)P\left(E_2\right)P\left(E_3\right)+P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)$
$=\frac{5}{7}\times \frac{4}{7}\times (1-\frac{3}{7})+\frac{5}{7}\times (1-\frac{4}{7})\times \frac{3}{7}(1-\frac{5}{7})\times \frac{4}{7}\times \frac{3}{7}+\frac{5}{7}\times \frac{4}{7}\times \frac{3}{7}$
$=\frac{209}{343}$