Question

Three critics review a book. Odds in favour of the book are 5 : 2, 4 : 3 and 3 : 4 respectively for the critics.
Find the probability that majority are in favour of the book.

• A. $\frac{108}{343}$
• B. $\frac{209}{343}$
• C. $\frac{1}{7}$
• D. $\frac{1}{243}$

Answer 1

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

Let A, B, C denote the events of favouring the book by the first, the second and the third critic respectively.
Then $P\left(A\right)=\frac{5}{7},P\left(B\right)=\frac{4}{7}andP\left(C\right)\frac{3}{7}$
$\therefore P\left(\overline{A}\right)=\frac{2}{7},P\left(\overline{B}\right)=\frac{3}{7}andP\left(\overline{C}\right)=\frac{4}{7}$
$\therefore$ Required probability
= P (Two favour the book or three favour the book)
= P (Two favour the book) + P (Three favour the book
= P [{A and B (not C)} or {A and (not B) and C} or {(not A) and B and C}] + P(A and B and C)
$=P\left[\right(A)\cap B\cap (\overline{C})\cup P(A\cap \overline{B}\cap C)\cup (\overline{A}\cap B\cap C)+P(A\cap B\cap C\left)\right]=P\left(A\right)P\left(B\right)P\left(\overline{C}\right)+P\left(A\right)P\left(\overline{B}\right)P\left(C\right)+p\left(\overline{A}\right)P\left(B\right)P\left(C\right)+P(A\cap B\cap C)=(\frac{5}{7}\times \frac{4}{7}\times \frac{4}{7})+(\frac{5}{7}\times \frac{3}{7}\times \frac{3}{7})+(\frac{2}{7}\times \frac{4}{7}\times \frac{3}{7})+(\frac{5}{7}\times \frac{4}{7}\times \frac{3}{7})=\frac{209}{343}$