A can hit a target 4 times in 5 shots B can hit 3 times in 4 shots and C can hit 2 times in 3 shots . The probability that B and C hit and A does not hit is

Options

A. `(1)/(10)`
B. `(2)/(5)`
C. `(7)/(10)`
D. none of these

Explanation

To solve the problem, we need to find the probability that B and C hit the target while A does not hit it. Let's break this down step by step:

### Step 1: Determine the probabilities of hitting the target for each person.
- **Probability that A hits the target (P(A))**: A can hit 4 times in 5 shots.
\[ P(A) = \frac{4}{5} \]
- **Probability that B hits the target (P(B))**: B can hit 3 times in 4 shots.
\[ P(B) = \frac{3}{4} \]
- **Probability that C hits the target (P(C))**: C can hit 2 times in 3 shots.
\[ P(C) = \frac{2}{3} \]

### Step 2: Determine the probability that A does not hit the target.
- **Probability that A does not hit the target (P(A'))**:
\[ P(A') = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5} \]

### Step 3: Calculate the combined probability that B and C hit the target while A does not.
Since the events are independent, we can multiply the probabilities of B hitting, C hitting, and A not hitting:
\[ P(B \text{ and } C \text{ and } A') = P(B) \times P(C) \times P(A') \]
Substituting the values we calculated:
\[ P(B \text{ and } C \text{ and } A') = P(B) \times P(C) \times P(A') = \left(\frac{3}{4}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{1}{5}\right) \]

### Step 4: Perform the multiplication.
Calculating the above expression:
\[ P(B \text{ and } C \text{ and } A') = \frac{3}{4} \times \frac{2}{3} \times \frac{1}{5} = \frac{3 \times 2 \times 1}{4 \times 3 \times 5} = \frac{6}{60} = \frac{1}{10} \]

### Final Answer:
The probability that B and C hit the target while A does not hit is:
\[ \frac{1}{10} \]