A,B and C can hit a target 3 times out of 5 trials, 4 times out of 5 trials and 2 times out of 5 trials. Find the probability that:
(i) exactly two can hit the target
(ii) at least two can hit the target.

Answer

To solve the problem, we need to find the probabilities for the two scenarios given the probabilities of A, B, and C hitting the target.

### Given:
- Probability of A hitting the target, \( P(A) = \frac{3}{5} \)
- Probability of B hitting the target, \( P(B) = \frac{4}{5} \)
- Probability of C hitting the target, \( P(C) = \frac{2}{5} \)

### Step 1: Calculate the probabilities of A, B, and C missing the target.
- Probability of A missing, \( P(A') = 1 - P(A) = 1 - \frac{3}{5} = \frac{2}{5} \)
- Probability of B missing, \( P(B') = 1 - P(B) = 1 - \frac{4}{5} = \frac{1}{5} \)
- Probability of C missing, \( P(C') = 1 - P(C) = 1 - \frac{2}{5} = \frac{3}{5} \)

### Part (i): Find the probability that exactly two can hit the target.

We need to consider the following cases where exactly two out of A, B, and C hit the target:

1. A hits, B hits, C misses: \( P(A) \cdot P(B) \cdot P(C') \)
2. A hits, B misses, C hits: \( P(A) \cdot P(B') \cdot P(C) \)
3. A misses, B hits, C hits: \( P(A') \cdot P(B) \cdot P(C) \)

Calculating each case:

1. \( P(A) \cdot P(B) \cdot P(C') = \frac{3}{5} \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{36}{125} \)
2. \( P(A) \cdot P(B') \cdot P(C) = \frac{3}{5} \cdot \frac{1}{5} \cdot \frac{2}{5} = \frac{6}{125} \)
3. \( P(A') \cdot P(B) \cdot P(C) = \frac{2}{5} \cdot \frac{4}{5} \cdot \frac{2}{5} = \frac{16}{125} \)

Now, we sum these probabilities:

\[ P(\text{exactly two hit}) = \frac{36}{125} + \frac{6}{125} + \frac{16}{125} = \frac{58}{125} \]

### Part (ii): Find the probability that at least two can hit the target.

At least two hitting means either exactly two hit or all three hit. We already calculated the probability for exactly two hitting. Now we need to calculate the probability for all three hitting:

- Probability that A, B, and C all hit: \( P(A) \cdot P(B) \cdot P(C) = \frac{3}{5} \cdot \frac{4}{5} \cdot \frac{2}{5} = \frac{24}{125} \)

Now, we sum the probabilities for at least two hitting:

\[ P(\text{at least two hit}) = P(\text{exactly two hit}) + P(\text{all three hit}) = \frac{58}{125} + \frac{24}{125} = \frac{82}{125} \]

### Final Answers:
(i) The probability that exactly two can hit the target is \( \frac{58}{125} \).

(ii) The probability that at least two can hit the target is \( \frac{82}{125} \).