Atul can hit a target 3 times in 6 shots, Bhola can hit the target 2 times in 6 shots and Chandra can hit the 4 times in 4 shots. What is the probability that at least 2 shots (out of 1 shot taken by each one of them) hit the target ?

Options

A. `1/2`
B. `2/3`
C. `1/3`
D. `5/6`

Answer

To solve the problem, we need to find the probability that at least 2 out of 3 shots (one taken by each person: Atul, Bhola, and Chandra) hit the target.

### Step-by-step Solution:

1. **Determine the individual probabilities of hitting the target:**
- **Atul:** Hits the target 3 times in 6 shots.
\[ P(A) = \frac{3}{6} = \frac{1}{2} \]
- **Bhola:** Hits the target 2 times in 6 shots.
\[ P(B) = \frac{2}{6} = \frac{1}{3} \]
- **Chandra:** Hits the target 4 times in 4 shots.
\[ P(C) = \frac{4}{4} = 1 \]

2. **Calculate the probabilities of missing the target:**
- **Atul:**
\[ P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \]
- **Bhola:**
\[ P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \]
- **Chandra:**
\[ P(C') = 1 - P(C) = 1 - 1 = 0 \]

3. **Calculate the probability of the complementary event (fewer than 2 hits):**
- The complementary event consists of two scenarios:
- Exactly 0 hits
- Exactly 1 hit

4. **Calculate the probability of exactly 0 hits:**
\[ P(0 \text{ hits}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times 0 = 0 \]

5. **Calculate the probability of exactly 1 hit:**
- There are three ways to have exactly 1 hit (Atul hits, Bhola hits, or Chandra hits):
- **Case 1:** Atul hits, Bhola and Chandra miss:
\[ P(A) \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times 0 = 0 \]
- **Case 2:** Bhola hits, Atul and Chandra miss:
\[ P(A') \times P(B) \times P(C') = \frac{1}{2} \times \frac{1}{3} \times 0 = 0 \]
- **Case 3:** Chandra hits, Atul and Bhola miss:
\[ P(A') \times P(B') \times P(C) = \frac{1}{2} \times \frac{2}{3} \times 1 = \frac{1}{3} \]

- Total probability of exactly 1 hit:
\[ P(1 \text{ hit}) = 0 + 0 + \frac{1}{3} = \frac{1}{3} \]

6. **Calculate the probability of at least 2 hits:**
\[ P(\text{at least 2 hits}) = 1 - P(0 \text{ hits}) - P(1 \text{ hit}) = 1 - 0 - \frac{1}{3} = \frac{2}{3} \]

### Final Answer:
The probability that at least 2 shots hit the target is \(\frac{2}{3}\).