To solve the problem, we need to calculate the probabilities based on the information given about A, B, and C's ability to hit 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, \( P(C) = \frac{2}{3} \)
### Step 1: Calculate the probability that A, B, and C all hit the target.
Since the events are independent, we can multiply their probabilities:
\[
P(A \text{ hits}) \times P(B \text{ hits}) \times P(C \text{ hits}) = P(A) \cdot P(B) \cdot P(C)
\]
Substituting the values:
\[
P(A \text{ hits}) = \frac{4}{5}, \quad P(B \text{ hits}) = \frac{3}{4}, \quad P(C \text{ hits}) = \frac{2}{3}
\]
Calculating:
\[
P(A \text{ hits}) \times P(B \text{ hits}) \times P(C \text{ hits}) = \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}
\]
Now, simplifying:
\[
= \frac{4 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{2}{5}
\]
### Answer for part (i):
The probability that A, B, and C all hit the target is \( \frac{2}{5} \).
---
### Step 2: Calculate the probability that B and C hit the target, but A does not hit the target.
First, we need to find the probability that A does not hit the target, which is the complement of A hitting:
\[
P(A \text{ does not hit}) = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5}
\]
Now, we need to find the probability that B and C hit the target:
\[
P(B \text{ hits}) \times P(C \text{ hits}) = P(B) \cdot P(C)
\]
Substituting the values:
\[
P(B \text{ hits}) = \frac{3}{4}, \quad P(C \text{ hits}) = \frac{2}{3}
\]
Calculating:
\[
P(B \text{ hits}) \times P(C \text{ hits}) = \frac{3}{4} \times \frac{2}{3}
\]
Now, simplifying:
\[
= \frac{3 \cdot 2}{4 \cdot 3} = \frac{1}{2}
\]
Now, combining both parts (A does not hit and B and C hit):
\[
P(A \text{ does not hit}) \times P(B \text{ hits}) \times P(C \text{ hits}) = P(A') \cdot P(B) \cdot P(C)
\]
Substituting the values:
\[
= \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3}
\]
Calculating:
\[
= \frac{1 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{6}{60} = \frac{1}{10}
\]
### Answer for part (ii):
The probability that B and C hit the target, but A does not hit the target is \( \frac{1}{10} \).
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