A can hit a target 4 times in 5 shots ,
B can hit a target 3 times in 4 shots ,
C can hit a target 2 times in 3 shots,
All the three fire a shot each. What is the probability that two shots are at least hit ?

Options

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

Explanation

To solve the problem, we need to calculate the probability that at least 2 shots are hit when A, B, and C each fire one shot at a target.

### Step-by-Step Solution:

1. **Determine the probabilities of hitting the target for A, B, and C:**
- 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, \( P(C) = \frac{2}{3} \)

2. **Calculate the probabilities of missing the target for A, B, and C:**
- Probability that A misses the target, \( P(A') = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5} \)
- Probability that B misses the target, \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)
- Probability that C misses the target, \( P(C') = 1 - P(C) = 1 - \frac{2}{3} = \frac{1}{3} \)

3. **Calculate the total probability of all possible outcomes:**
- The total number of outcomes when A, B, and C shoot is \( 2^3 = 8 \) (each can either hit or miss).

4. **Find the probabilities for the scenarios where at least 2 shots are hit:**
- We can find the probability of the complementary event (0 or 1 hit) and subtract it from 1.

5. **Calculate the probability of 0 hits:**
\[ P(0 \text{ hits}) = P(A') \cdot P(B') \cdot P(C') = \frac{1}{5} \cdot \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{60} \]

6. **Calculate the probability of exactly 1 hit:**
- There are 3 ways to have exactly 1 hit (A hits, B misses, C misses; A misses, B hits, C misses; A misses, B misses, C hits):
\[ P(1 \text{ hit}) = P(A) \cdot P(B') \cdot P(C') + P(A') \cdot P(B) \cdot P(C') + P(A') \cdot P(B') \cdot P(C) \]
\[ = \left(\frac{4}{5} \cdot \frac{1}{4} \cdot \frac{1}{3}\right) + \left(\frac{1}{5} \cdot \frac{3}{4} \cdot \frac{1}{3}\right) + \left(\frac{1}{5} \cdot \frac{1}{4} \cdot \frac{2}{3}\right) \]
\[ = \frac{4}{60} + \frac{3}{60} + \frac{2}{60} = \frac{9}{60} = \frac{3}{20} \]

7. **Calculate the probability of at least 2 hits:**
\[ P(\text{at least 2 hits}) = 1 - P(0 \text{ hits}) - P(1 \text{ hit}) \]
\[ = 1 - \frac{1}{60} - \frac{3}{20} \]
\[ = 1 - \frac{1}{60} - \frac{9}{60} = 1 - \frac{10}{60} = 1 - \frac{1}{6} = \frac{5}{6} \]

### Final Answer:
The probability that at least 2 shots are hit is \( \frac{5}{6} \).