Three candidates A, B, and C are going to play in a chess competition to win FIDE (World Chess Federation) cup this year. A is thrice as likely to win as B and B is twice as likely as to win C. Find the respective probability of A, B and C to win the cup.

Answer

To solve the problem, we need to determine the respective probabilities of candidates A, B, and C winning the chess competition based on their likelihoods of winning.

### Step-by-Step Solution:

**Step 1: Define the Likelihoods**
Let the likelihood of C winning be represented as \( k \). According to the problem:
- B is twice as likely to win as C, so the likelihood of B winning is \( 2k \).
- A is thrice as likely to win as B, so the likelihood of A winning is \( 3 \times (2k) = 6k \).

**Step 2: Calculate Total Likelihood**
Now, we sum up the likelihoods of all candidates:
\[ \text{Total likelihood} = \text{Likelihood of A} + \text{Likelihood of B} + \text{Likelihood of C} = 6k + 2k + k = 9k. \]

**Step 3: Calculate Individual Probabilities**
Now we can calculate the probability of each candidate winning by dividing their respective likelihood by the total likelihood.

- **Probability of A winning**:
\[ P(A) = \frac{\text{Likelihood of A}}{\text{Total likelihood}} = \frac{6k}{9k} = \frac{6}{9} = \frac{2}{3}. \]

- **Probability of B winning**:
\[ P(B) = \frac{\text{Likelihood of B}}{\text{Total likelihood}} = \frac{2k}{9k} = \frac{2}{9}. \]

- **Probability of C winning**:
\[ P(C) = \frac{\text{Likelihood of C}}{\text{Total likelihood}} = \frac{k}{9k} = \frac{1}{9}. \]

**Step 4: Final Probabilities**
Thus, the respective probabilities of A, B, and C winning the cup are:
- \( P(A) = \frac{2}{3} \)
- \( P(B) = \frac{2}{9} \)
- \( P(C) = \frac{1}{9} \)

### Summary of Results:
- Probability of A winning: \( \frac{2}{3} \)
- Probability of B winning: \( \frac{2}{9} \)
- Probability of C winning: \( \frac{1}{9} \)