Question

The odds that a book will be reviewed favourably by three independent critics are 5 to 2,4 to 3 and 3 to 4 respectively; what is the probability that of three reviews a majority will be favourable?

• A. $p=\frac{149}{343}$
• B. $p=\frac{209}{343}$
• C. $p=\frac{129}{343}$
• D. $p=\frac{185}{343}$

Answer 1

The correct option is B $p=\frac{209}{343}$

Let $p_1,p_2,p_3$ denote the probabilities that first, second and third critics review the book.
Given odds in favour that first critic will review the book is $5:2$
$p_1=\frac{5}{5+2}=\frac{5}{7}$
$\Rightarrow \overline{p_1}=1-\frac{5}{7}=\frac{2}{7}$
Given odds in favour that second critic will review the book is $4:3$
$p_2=\frac{4}{4+3}=\frac{4}{7}$
$\Rightarrow \overline{p_2}=1-\frac{4}{7}=\frac{3}{7}$
Given odds in favour that third critic will review the book is $3:4$
$p_3=\frac{3}{3+4}=\frac{3}{7}$
$\Rightarrow \overline{p_3}=1-\frac{3}{7}=\frac{4}{7}$
Now the majority will be favourable if any of the two critics is favourable and third is unfavourable or all the three critics are favourable.
Hence the required probability is
$=p_1{p}_2\overline{p_3}+p_1\overline{p_2}{p}_3+\overline{p_1}{p}_2{p}_3+p_1{p}_2{p}_3$

$=\frac{5}{7}\cdot \frac{4}{7}\cdot \frac{4}{7}+\frac{5}{7}\cdot \frac{3}{7}\cdot \frac{3}{7}+\frac{2}{7}\cdot \frac{4}{7}\cdot \frac{3}{7}+\frac{5}{7}\cdot \frac{4}{7}\cdot \frac{3}{7}$
$=\frac{209}{343}$