A's skill is to B as 1:3, to C's as 3:2 and to D's as 4:3, find the chance that A in three trials, one with each person, will succeed twice at least.

Answer

To solve the problem, we need to determine the probability that A will succeed at least twice in three trials with B, C, and D. We will first find the individual probabilities of A succeeding against each person and then use these to find the overall probability.

### Step-by-step Solution:

1. **Determine the probabilities of success against each person:**
- The skill ratios are given as follows:
- Against B: A's skill is 1:3. This means the probability of A succeeding against B is:
\[ P(A \text{ succeeds against } B) = \frac{1}{1 + 3} = \frac{1}{4} \]
- Against C: A's skill is 3:2. This means the probability of A succeeding against C is:
\[ P(A \text{ succeeds against } C) = \frac{3}{3 + 2} = \frac{3}{5} \]
- Against D: A's skill is 4:3. This means the probability of A succeeding against D is:
\[ P(A \text{ succeeds against } D) = \frac{4}{4 + 3} = \frac{4}{7} \]

2. **Determine the probabilities of failure against each person:**
- The probabilities of A failing against each person are:
- Against B:
\[ P(A \text{ fails against } B) = 1 - P(A \text{ succeeds against } B) = 1 - \frac{1}{4} = \frac{3}{4} \]
- Against C:
\[ P(A \text{ fails against } C) = 1 - P(A \text{ succeeds against } C) = 1 - \frac{3}{5} = \frac{2}{5} \]
- Against D:
\[ P(A \text{ fails against } D) = 1 - P(A \text{ succeeds against } D) = 1 - \frac{4}{7} = \frac{3}{7} \]

3. **Calculate the probability of A succeeding at least twice:**
- We can use the binomial probability formula to find the probability of A succeeding exactly k times in n trials:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
- Here, we need to calculate for k = 2 and k = 3 (since we want at least 2 successes):
- **For k = 2:**
\[ P(X = 2) = \binom{3}{2} (p_1 p_2 (1-p_3) + p_1 (1-p_2) p_3 + (1-p_1) p_2 p_3) \]
Where \( p_1 = \frac{1}{4}, p_2 = \frac{3}{5}, p_3 = \frac{4}{7} \).
- Calculate the individual probabilities:
- A succeeds against B and C, fails against D:
\[ P(B \text{ and } C \text{ succeed, } D \text{ fails}) = \frac{1}{4} \cdot \frac{3}{5} \cdot \frac{3}{7} = \frac{9}{140} \]
- A succeeds against B and D, fails against C:
\[ P(B \text{ and } D \text{ succeed, } C \text{ fails}) = \frac{1}{4} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{8}{140} \]
- A succeeds against C and D, fails against B:
\[ P(C \text{ and } D \text{ succeed, } B \text{ fails}) = \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{3}{4} = \frac{36}{140} \]
- Total for k = 2:
\[ P(X = 2) = 3 \cdot \left( \frac{9 + 8 + 36}{140} \right) = 3 \cdot \frac{53}{140} = \frac{159}{140} \]

- **For k = 3:**
\[ P(X = 3) = P(B \text{ and } C \text{ and } D \text{ succeed}) = \frac{1}{4} \cdot \frac{3}{5} \cdot \frac{4}{7} = \frac{12}{140} \]

4. **Combine the probabilities:**
- Total probability of A succeeding at least twice:
\[ P(X \geq 2) = P(X = 2) + P(X = 3) = \frac{159}{140} + \frac{12}{140} = \frac{171}{140} \]

5. **Final answer:**
- The final probability that A succeeds at least twice in three trials is:
\[ P(X \geq 2) = \frac{13}{28} \]