Three critics review a book. For the three critics the odds in favor of the book are $2: 5$, $3: 4$ and $4:…
Three critics review a book. For the three critics the odds in favor of the book are $2: 5$, $3: 4$ and $4: 3$ respectively. The probability that the majority is in favor of the book, is given by
$\frac{183}{343}$
$\frac{160}{343}$
$\frac{209}{343}$
$\frac{134}{343}$
Solution
The probability that the first critic favors the book is $\mathrm{P}(\mathrm{A})=\frac{2}{2+5}=\frac{2}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\frac{2}{7}=\frac{5}{7}$
The probability that the second critic favors the book is $\mathrm{P}(\mathrm{B})=\frac{3}{3+4}=\frac{3}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{3}{7}=\frac{4}{7}$
The probability that the third critic favors the book is $\mathrm{P}(\mathrm{C})=\frac{4}{4+3}=\frac{4}{7}$
$\therefore \quad \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{4}{7}=\frac{3}{7}$
$\therefore \quad$ Majority will be in favor of the book if at least two critics favor the book.
Hence, the probability is
$\begin{aligned}
& \mathrm{P}\left(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}^{\prime}\right)+\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}\right) \\
& \quad+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B} \cap \mathrm{C}\right)+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \\
& =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}(\mathrm{C}) \\
& \quad+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C}) \\
& =\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7} \\
& =\frac{18}{343}+\frac{32}{343}+\frac{60}{343}+\frac{24}{343}=\frac{134}{343}
\end{aligned}$