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`

Answer

To solve the problem, we need to calculate the probability that at least two out of the three shooters (A, B, and C) hit the target when they each take one shot.

### Step 1: Determine individual probabilities of hitting and missing the target.

- **For Shooter A:**
- Probability of hitting the target (P(A)) = 4/5
- Probability of missing the target (P(A')) = 1 - P(A) = 1/5

- **For Shooter B:**
- Probability of hitting the target (P(B)) = 3/4
- Probability of missing the target (P(B')) = 1 - P(B) = 1/4

- **For Shooter C:**
- Probability of hitting the target (P(C)) = 2/3
- Probability of missing the target (P(C')) = 1 - P(C) = 1/3

### Step 2: Calculate the probabilities for different cases where at least two shooters hit the target.

We can break this down into the following cases:

1. **All three hit the target (A, B, C):**
\[ P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \]
Simplifying:
\[ = \frac{4 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{2}{5} \]

2. **A misses, B and C hit (A', B, C):**
\[ P(A' \cap B \cap C) = P(A') \cdot P(B) \cdot P(C) = \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \]
Simplifying:
\[ = \frac{1 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{1}{10} \]

3. **A hits, B misses, C hits (A, B', C):**
\[ P(A \cap B' \cap C) = P(A) \cdot P(B') \cdot P(C) = \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} \]
Simplifying:
\[ = \frac{4 \cdot 1 \cdot 2}{5 \cdot 4 \cdot 3} = \frac{2}{15} \]

4. **A hits, B hits, C misses (A, B, C'):**
\[ P(A \cap B \cap C') = P(A) \cdot P(B) \cdot P(C') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \]
Simplifying:
\[ = \frac{4 \cdot 3 \cdot 1}{5 \cdot 4 \cdot 3} = \frac{1}{5} \]

### Step 3: Add the probabilities of all cases where at least two shooters hit the target.

Now, we sum the probabilities from all cases:
\[ P(\text{at least 2 hits}) = P(A \cap B \cap C) + P(A' \cap B \cap C) + P(A \cap B' \cap C) + P(A \cap B \cap C') \]
Substituting the values:
\[ = \frac{2}{5} + \frac{1}{10} + \frac{2}{15} + \frac{1}{5} \]

### Step 4: Find a common denominator and sum the fractions.

The least common multiple (LCM) of 5, 10, and 15 is 30. We convert each fraction:
- \(\frac{2}{5} = \frac{12}{30}\)
- \(\frac{1}{10} = \frac{3}{30}\)
- \(\frac{2}{15} = \frac{4}{30}\)
- \(\frac{1}{5} = \frac{6}{30}\)

Now we add them:
\[ P(\text{at least 2 hits}) = \frac{12}{30} + \frac{3}{30} + \frac{4}{30} + \frac{6}{30} = \frac{25}{30} \]

### Step 5: Simplify the final probability.

\[ P(\text{at least 2 hits}) = \frac{25}{30} = \frac{5}{6} \]

Thus, the probability that at least two shots hit the target is \(\frac{5}{6}\).