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

Options

A. `35//49`
B. `125//343`
C. `164//343`
D. `209//343`

Explanation

To solve the problem step by step, we need to calculate the probability that at least two out of three critics favor the book.

### Step 1: Calculate the probabilities for each critic

1. **First critic**: The odds in favor are 5:2.
- Probability \( P(E_1) = \frac{5}{5 + 2} = \frac{5}{7} \)

2. **Second critic**: The odds in favor are 4:3.
- Probability \( P(E_2) = \frac{4}{4 + 3} = \frac{4}{7} \)

3. **Third critic**: The odds in favor are 3:4.
- Probability \( P(E_3) = \frac{3}{3 + 4} = \frac{3}{7} \)

### Step 2: Calculate the probabilities of the critics not favoring the book

1. **First critic not favoring**:
- Probability \( P(E_1') = 1 - P(E_1) = 1 - \frac{5}{7} = \frac{2}{7} \)

2. **Second critic not favoring**:
- Probability \( P(E_2') = 1 - P(E_2) = 1 - \frac{4}{7} = \frac{3}{7} \)

3. **Third critic not favoring**:
- Probability \( P(E_3') = 1 - P(E_3) = 1 - \frac{3}{7} = \frac{4}{7} \)

### Step 3: Calculate the probability of at least 2 critics favoring the book

To find the probability that at least 2 critics favor the book, we can consider the following scenarios:

1. **Exactly 2 critics favor the book**:
- \( P(E_1 \cap E_2 \cap E_3') + P(E_1 \cap E_2' \cap E_3) + P(E_1' \cap E_2 \cap E_3) \)

- **Calculating each term**:
- \( P(E_1 \cap E_2 \cap E_3') = P(E_1) \cdot P(E_2) \cdot P(E_3') = \frac{5}{7} \cdot \frac{4}{7} \cdot \frac{4}{7} = \frac{80}{343} \)
- \( P(E_1 \cap E_2' \cap E_3) = P(E_1) \cdot P(E_2') \cdot P(E_3) = \frac{5}{7} \cdot \frac{3}{7} \cdot \frac{3}{7} = \frac{45}{343} \)
- \( P(E_1' \cap E_2 \cap E_3) = P(E_1') \cdot P(E_2) \cdot P(E_3) = \frac{2}{7} \cdot \frac{4}{7} \cdot \frac{3}{7} = \frac{24}{343} \)

- **Total for exactly 2 critics**:
- \( P(\text{Exactly 2}) = \frac{80}{343} + \frac{45}{343} + \frac{24}{343} = \frac{149}{343} \)

2. **All 3 critics favor the book**:
- \( P(E_1 \cap E_2 \cap E_3) = P(E_1) \cdot P(E_2) \cdot P(E_3) = \frac{5}{7} \cdot \frac{4}{7} \cdot \frac{3}{7} = \frac{60}{343} \)

### Step 4: Combine the probabilities

Now, we can find the total probability that at least 2 critics favor the book:

\[ P(\text{At least 2}) = P(\text{Exactly 2}) + P(\text{All 3}) = \frac{149}{343} + \frac{60}{343} = \frac{209}{343} \]

### Final Answer

The probability that the majority of critics favor the book is \( \frac{209}{343} \).

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