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(D) The probability that the first critic favours the book is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.$	herefore P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.

Vedclass · Accepted Answer

$\frac{134}{343}$

Vedclass · Answer

(D) The probability that the first critic favours the book is $P(A) = \frac{2}{2+5} = \frac{2}{7}$.$	herefore P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.The probability that the second critic favours the book is $P(B) = \frac{3}{3+4} = \frac{3}{7}$.$	herefore P(B') = 1 - \frac{3}{7} = \frac{4}{7}$.The probability that the third critic favours the book is $P(C) = \frac{4}{4+3} = \frac{4}{7}$.$	herefore P(C') = 1 - \frac{4}{7} = \frac{3}{7}$.Majority will be in favour of the book if at least two critics favour the book.Hence,the probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$.$= P(A) \cdot P(B) \cdot P(C') + P(A) \cdot P(B') \cdot P(C) + P(A') \cdot P(B) \cdot P(C) + P(A) \cdot P(B) \cdot P(C)$.$= \left(\frac{2}{7} 	imes \frac{3}{7} 	imes \frac{3}{7}ight) + \left(\frac{2}{7} 	imes \frac{4}{7} 	imes \frac{4}{7}ight) + \left(\frac{5}{7} 	imes \frac{3}{7} 	imes \frac{4}{7}ight) + \left(\frac{2}{7} 	imes \frac{3}{7} 	imes \frac{4}{7}ight)$.$= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$0.5$	$0.35$	$.........$	$0.7$

Three critics review a book. For the three critics,the odds in favour of the book are $2:5$,$3:4$,and $4:3$ respectively. The probability that the majority is in favour of the book is given by

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