A can hit a target 3 times out of 7,B,4 times out of 5, C,2 times out of 3, They all fire together. What is the probability tha tat least two shots hit the target.

Answer

To solve the problem, we need to find the probability that at least two out of three shooters (A, B, and C) hit a target when they fire simultaneously. Let's denote the probabilities of hitting the target for each shooter:

- Probability that A hits the target, \( P(A) = \frac{3}{7} \)
- Probability that B hits the target, \( P(B) = \frac{4}{5} \)
- Probability that C hits the target, \( P(C) = \frac{2}{3} \)

The probability that a shooter misses the target is given by \( P(\text{miss}) = 1 - P(\text{hit}) \):

- Probability that A misses the target, \( P(A') = 1 - P(A) = 1 - \frac{3}{7} = \frac{4}{7} \)
- Probability that B misses the target, \( P(B') = 1 - P(B) = 1 - \frac{4}{5} = \frac{1}{5} \)
- Probability that C misses the target, \( P(C') = 1 - P(C) = 1 - \frac{2}{3} = \frac{1}{3} \)

We need to calculate the probability of at least two hits. This can be calculated using the complement rule:

\[ P(\text{at least 2 hits}) = 1 - P(\text{0 hits}) - P(\text{1 hit}) \]

### Step 1: Calculate \( P(0 \text{ hits}) \)

The probability that none of the shooters hit the target is:

\[ P(0 \text{ hits}) = P(A') \cdot P(B') \cdot P(C') = \frac{4}{7} \cdot \frac{1}{5} \cdot \frac{1}{3} \]

Calculating this:

\[ P(0 \text{ hits}) = \frac{4}{7} \cdot \frac{1}{5} \cdot \frac{1}{3} = \frac{4}{105} \]

### Step 2: Calculate \( P(1 \text{ hit}) \)

To find the probability of exactly one hit, we consider the cases where each shooter hits while the others miss:

1. A hits, B misses, C misses:
\[ P(A) \cdot P(B') \cdot P(C') = \frac{3}{7} \cdot \frac{1}{5} \cdot \frac{1}{3} = \frac{3}{105} \]

2. A misses, B hits, C misses:
\[ P(A') \cdot P(B) \cdot P(C') = \frac{4}{7} \cdot \frac{4}{5} \cdot \frac{1}{3} = \frac{16}{105} \]

3. A misses, B misses, C hits:
\[ P(A') \cdot P(B') \cdot P(C) = \frac{4}{7} \cdot \frac{1}{5} \cdot \frac{2}{3} = \frac{8}{105} \]

Now, summing these probabilities gives us \( P(1 \text{ hit}) \):

\[ P(1 \text{ hit}) = \frac{3}{105} + \frac{16}{105} + \frac{8}{105} = \frac{27}{105} \]

### Step 3: Calculate \( P(\text{at least 2 hits}) \)

Now we can find the probability of at least two hits:

\[ P(\text{at least 2 hits}) = 1 - P(0 \text{ hits}) - P(1 \text{ hit}) \]

Substituting the values we calculated:

\[ P(\text{at least 2 hits}) = 1 - \frac{4}{105} - \frac{27}{105} = 1 - \frac{31}{105} = \frac{74}{105} \]

### Final Answer

Thus, the probability that at least two shots hit the target is:

\[ \boxed{\frac{74}{105}} \]