Question

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

• A. $\frac{183}{343}$
• B. $\frac{160}{343}$
• C. $\frac{209}{343}$
• D. $\frac{134}{343}$

Hint:

Probability Tip: When given "odds in favor" as $a:b$, the probability of the event occurring is always $\frac{a}{a+b}$. If it were "odds against", it would be $\frac{b}{a+b}$.

Answer 1

The Correct Option is D

Concept: Probability - Independent Events and Odds.

Step 1: Determine the probability of each critic favoring the book. The odds in favor of an event are given as $a:b$, meaning the probability of the event occurring is $P = \frac{a}{a+b}$. For the first critic (let's call it event A), the odds are 2:5. So, $P(A) = \frac{2}{2+5} = \frac{2}{7}$, and the probability of not favoring is $P(A') = 1 - \frac{2}{7} = \frac{5}{7}$.

Step 2: Calculate the probabilities for the second and third critics. For the second critic (event B), the odds are 3:4. Thus, $P(B) = \frac{3}{3+4} = \frac{3}{7}$, and $P(B') = \frac{4}{7}$. For the third critic (event C), the odds are 4:3. Thus, $P(C) = \frac{4}{4+3} = \frac{4}{7}$, and $P(C') = \frac{3}{7}$.

Step 3: Define the condition for the "majority". A "majority" out of three critics means that at least two critics are in favor of the book. This can happen in four distinct ways: 1) A and B are in favor, C is not ($A \cap B \cap C'$). 2) A and C are in favor, B is not ($A \cap B' \cap C$). 3) B and C are in favor, A is not ($A' \cap B \cap C$). 4) All three are in favor ($A \cap B \cap C$).

Step 4: Calculate the probability of each individual case. Since the critics' reviews are independent, we multiply their individual probabilities: $P(A \cap B \cap C') = P(A) \cdot P(B) \cdot P(C') = \left(\frac{2}{7}\right) \left(\frac{3}{7}\right) \left(\frac{3}{7}\right) = \frac{18}{343}$. $P(A \cap B' \cap C) = P(A) \cdot P(B') \cdot P(C) = \left(\frac{2}{7}\right) \left(\frac{4}{7}\right) \left(\frac{4}{7}\right) = \frac{32}{343}$. $P(A' \cap B \cap C) = P(A') \cdot P(B) \cdot P(C) = \left(\frac{5}{7}\right) \left(\frac{3}{7}\right) \left(\frac{4}{7}\right) = \frac{60}{343}$. $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) = \left(\frac{2}{7}\right) \left(\frac{3}{7}\right) \left(\frac{4}{7}\right) = \frac{24}{343}$.

Step 5: Sum the probabilities to find the final result. Add the probabilities of all four mutually exclusive cases: Total Probability = $\frac{18}{343} + \frac{32}{343} + \frac{60}{343} + \frac{24}{343} = \frac{18 + 32 + 60 + 24}{343} = \frac{134}{343}$. $$ \therefore \text{The probability that the majority is in favour is } \frac{134}{343}. $$