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?

Engineering

Mathematics

Conditional Probability

Question

$p = \frac{149}{343}$

$p = \frac{209}{343}$

$p = \frac{129}{343}$

$p = \frac{185}{343}$

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Let P₁, P₂, P₃ 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
$p_{1} = \frac{5}{5 + 2} = \frac{5}{7}$
${\bar{p}}_{1} = 1 - \frac{5}{7} = \frac{2}{7}$
Given odds in favour that second critic will review the book is 4:3
$p_{2} = \frac{4}{4 + 3} = \frac{4}{7}$
${\bar{p}}_{2} = 1 - \frac{4}{7} = \frac{3}{7}$
Given odds in favour that third critic will review the book is 3:4
$p_{3} = \frac{3}{3 + 4} = \frac{3}{7}$
${\bar{p}}_{3} = 1 - \frac{3}{7} = \frac{4}{7}$
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
$= P 1 P 2 P 3 +P 1 P 2 P 3 + P 1 P 2 P 3 + P 1 P 2 P 3$
$= \frac{5}{7} \cdot \frac{4}{7} \cdot \frac{4}{7} + \frac{5}{7} \cdot \frac{3}{7} \cdot \frac{3}{7} + \frac{2}{7} \cdot \frac{4}{7} \cdot \frac{3}{7} + \frac{5}{7} \cdot \frac{4}{7} \cdot \frac{3}{7}$
$= \frac{209}{343}$

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