Three persons A,B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2, respectively. The probability of two hits is

Options

A. 0.024
B. 0.188
C. 0.336
D. 0.452

Explanation

To solve the problem of finding the probability of exactly two hits when persons A, B, and C fire at a target in turn, we can follow these steps:

### Step 1: Identify the probabilities
- Probability that A hits the target, \( P(A) = 0.4 \)
- Probability that B hits the target, \( P(B) = 0.3 \)
- Probability that C hits the target, \( P(C) = 0.2 \)

The probabilities of missing the target are:
- Probability that A misses, \( P(A') = 1 - P(A) = 0.6 \)
- Probability that B misses, \( P(B') = 1 - P(B) = 0.7 \)
- Probability that C misses, \( P(C') = 1 - P(C) = 0.8 \)

### Step 2: Identify the combinations for exactly two hits
The possible combinations for exactly two hits (denoted as H for hit and M for miss) are:
1. A hits, B hits, C misses (A, B, C')
2. A hits, B misses, C hits (A, B', C)
3. A misses, B hits, C hits (A', B, C)

### Step 3: Calculate the probability for each combination
1. **For combination (A, B, C')**:
\[ P(A \text{ hits}) \times P(B \text{ hits}) \times P(C \text{ misses}) = P(A) \times P(B) \times P(C') = 0.4 \times 0.3 \times 0.8 = 0.096 \]

2. **For combination (A, B', C)**:
\[ P(A \text{ hits}) \times P(B \text{ misses}) \times P(C \text{ hits}) = P(A) \times P(B') \times P(C) = 0.4 \times 0.7 \times 0.2 = 0.056 \]

3. **For combination (A', B, C)**:
\[ P(A \text{ misses}) \times P(B \text{ hits}) \times P(C \text{ hits}) = P(A') \times P(B) \times P(C) = 0.6 \times 0.3 \times 0.2 = 0.036 \]

### Step 4: Sum the probabilities of all combinations
Now, we sum the probabilities of all three combinations to find the total probability of exactly two hits:
\[ P(\text{exactly 2 hits}) = P(A, B, C') + P(A, B', C) + P(A', B, C) = 0.096 + 0.056 + 0.036 = 0.188 \]

### Final Answer
The probability of exactly two hits is \( \boxed{0.188} \).