To solve the problem, we need to calculate the probabilities for two scenarios involving three persons A, B, and C shooting at a target.
### Given:
- Probability that A hits the target, \( P(A) = \frac{4}{5} \)
- Probability that B hits the target, \( P(B) = \frac{3}{4} \)
- Probability that C hits the target, \( P(C) = \frac{2}{3} \)
From these, we can find the probabilities that each person does not hit the target:
- Probability that A does not hit the target, \( P(A') = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5} \)
- Probability that B does not hit the target, \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)
- Probability that C does not hit the target, \( P(C') = 1 - P(C) = 1 - \frac{2}{3} = \frac{1}{3} \)
### (i) Probability that exactly two persons hit the target
To find the probability that exactly two persons hit the target, we can consider the following cases:
1. A and B hit the target, but C does not.
2. A and C hit the target, but B does not.
3. B and C hit the target, but A does not.
We can calculate the probability for each case:
1. **Case 1: A and B hit, C does not**
\[
P(A \cap B \cap C') = P(A) \cdot P(B) \cdot P(C') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3}
\]
\[
= \frac{4 \cdot 3 \cdot 1}{5 \cdot 4 \cdot 3} = \frac{4}{60} = \frac{1}{15}
\]
2. **Case 2: A and C hit, B does not**
\[
P(A \cap B' \cap C) = P(A) \cdot P(B') \cdot P(C) = \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3}
\]
\[
= \frac{4 \cdot 1 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{8}{60} = \frac{2}{15}
\]
3. **Case 3: B and C hit, A does not**
\[
P(A' \cap B \cap C) = P(A') \cdot P(B) \cdot P(C) = \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}
\]
\[
= \frac{1 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{6}{60} = \frac{1}{10}
\]
Now, we sum up the probabilities of all three cases:
\[
P(\text{exactly two hit}) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C)
\]
\[
= \frac{1}{15} + \frac{2}{15} + \frac{1}{10}
\]
To add these fractions, we need a common denominator. The least common multiple of 15 and 10 is 30:
\[
= \frac{2}{30} + \frac{4}{30} + \frac{3}{30} = \frac{9}{30} = \frac{3}{10}
\]
### (ii) Probability that none hit the target
The probability that none of them hit the target is given by:
\[
P(A') \cdot P(B') \cdot P(C') = \frac{1}{5} \cdot \frac{1}{4} \cdot \frac{1}{3}
\]
\[
= \frac{1 \cdot 1 \cdot 1}{5 \cdot 4 \cdot 3} = \frac{1}{60}
\]
### Final Answers:
1. The probability that exactly two persons hit the target is \( \frac{3}{10} \).
2. The probability that none hit the target is \( \frac{1}{60} \).
