Three critics review a book. For the three cr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $A, B, C$ be the events that the first,second,and third critics favor the book respectively. The probabilities are: $P(A) = \frac{2}{2+5} = \f

Vedclass · Accepted Answer

$\frac{134}{343}$

Vedclass · Answer

(D) Let $A, B, C$ be the events that the first,second,and third critics favor the book respectively. The probabilities are: $P(A) = \frac{2}{2+5} = \frac{2}{7}, P(A') = \frac{5}{7}$ $P(B) = \frac{3}{3+4} = \frac{3}{7}, P(B') = \frac{4}{7}$ $P(C) = \frac{4}{4+3} = \frac{4}{7}, P(C') = \frac{3}{7}$ The majority is in favor if at least two critics favor the book. The probability is $P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C)$. $= (\frac{2}{7} 	imes \frac{3}{7} 	imes \frac{3}{7}) + (\frac{2}{7} 	imes \frac{4}{7} 	imes \frac{4}{7}) + (\frac{5}{7} 	imes \frac{3}{7} 	imes \frac{4}{7}) + (\frac{2}{7} 	imes \frac{3}{7} 	imes \frac{4}{7})$ $= \frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{134}{343}$.

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

Three critics review a book. For the three critics,the odds in 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

Similar Questions

Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random,then the probability of the number being divisible by $2$ or $3$ is:

The odds in favour of drawing a king from a pack of $52$ playing cards is

The probability that a leap year selected at random contains either $53$ Sundays or $53$ Mondays is:

Fill in the blanks in the following table:

$P(A)$ $P(B)$ $P(A \cap B)$ $P(A \cup B)$

$0.5$ $0.35$ $.........$ $0.7$

If $A$ and $B$ are two events in a random experiment such that $P(A) + P(B) = 2 P(A \cap B)$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module