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.

Explanation

To solve the problem, we need to determine the probabilities for A, B, and C hitting the target, and then use these probabilities to find the required probabilities for the two scenarios.

### Step 1: Determine the probabilities of hitting the target for A, B, and C.

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

- Probability that B hits the target, \( P(B) = \frac{4}{5} \)
- Probability that B misses the target, \( P(B') = 1 - P(B) = 1 - \frac{4}{5} = \frac{1}{5} \)

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

### Step 2: Calculate the probability that exactly two can hit the target.

To find the probability that exactly two out of A, B, and C hit the target, we consider the following combinations:

1. A hits, B hits, C misses.
2. A hits, C hits, B misses.
3. B hits, C hits, A misses.

We will calculate the probability for each case and then sum them up.

#### Case 1: A hits, B hits, C misses
\[ P(A \text{ hits}, B \text{ hits}, C \text{ misses}) = P(A) \cdot P(B) \cdot P(C') \]
\[ = \frac{3}{5} \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{36}{125} \]

#### Case 2: A hits, C hits, B misses
\[ P(A \text{ hits}, C \text{ hits}, B \text{ misses}) = P(A) \cdot P(C) \cdot P(B') \]
\[ = \frac{3}{5} \cdot \frac{2}{5} \cdot \frac{1}{5} = \frac{6}{125} \]

#### Case 3: B hits, C hits, A misses
\[ P(B \text{ hits}, C \text{ hits}, A \text{ misses}) = P(B) \cdot P(C) \cdot P(A') \]
\[ = \frac{4}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{16}{125} \]

### Step 3: Sum the probabilities of the three cases
\[ P(\text{exactly 2 hit}) = \frac{36}{125} + \frac{6}{125} + \frac{16}{125} = \frac{58}{125} \]

### Step 4: Calculate the probability that at least two can hit the target.

To find the probability that at least two can hit the target, we can use the complement rule. We can find the probability that fewer than two hit the target (i.e., either none or only one hits) and subtract it from 1.

#### Case 1: None hit
\[ P(A') \cdot P(B') \cdot P(C') = \frac{2}{5} \cdot \frac{1}{5} \cdot \frac{3}{5} = \frac{6}{125} \]

#### Case 2: Exactly one hits
1. A hits, B misses, C misses:
\[ P(A) \cdot P(B') \cdot P(C') = \frac{3}{5} \cdot \frac{1}{5} \cdot \frac{3}{5} = \frac{9}{125} \]

2. B hits, A misses, C misses:
\[ P(B) \cdot P(A') \cdot P(C') = \frac{4}{5} \cdot \frac{2}{5} \cdot \frac{3}{5} = \frac{24}{125} \]

3. C hits, A misses, B misses:
\[ P(C) \cdot P(A') \cdot P(B') = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{1}{5} = \frac{4}{125} \]

#### Sum the probabilities for exactly one hitting:
\[ P(\text{exactly 1 hit}) = \frac{9}{125} + \frac{24}{125} + \frac{4}{125} = \frac{37}{125} \]

### Step 5: Calculate the probability of at least two hitting
\[ P(\text{at least 2 hit}) = 1 - P(\text{fewer than 2 hit}) \]
\[ = 1 - \left( P(\text{none hit}) + P(\text{exactly 1 hit}) \right) \]
\[ = 1 - \left( \frac{6}{125} + \frac{37}{125} \right) \]
\[ = 1 - \frac{43}{125} = \frac{82}{125} \]

### 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 \( \frac{82}{125} \).