The probability that a bomb dropped from a plane strikes the target is 1/5. the probability that out of six bombs dropped, at least two bombs strike the target is ____

Answer

To solve the problem, we need to find the probability that at least 2 bombs strike the target when 6 bombs are dropped, given that the probability of a single bomb hitting the target is \( \frac{1}{5} \).

### Step-by-Step Solution:

1. **Define the Probability of Hitting and Missing:**
- Let \( p = \frac{1}{5} \) (probability of hitting the target).
- Let \( q = 1 - p = \frac{4}{5} \) (probability of missing the target).

2. **Identify the Total Number of Trials:**
- We have \( n = 6 \) bombs dropped.

3. **Use the Complement Rule:**
- We need to find the probability of at least 2 bombs hitting the target. This can be calculated using the complement:
\[ P(\text{at least 2 hits}) = 1 - P(\text{0 hits}) - P(\text{1 hit}) \]

4. **Calculate \( P(0 \text{ hits}) \):**
- The probability that none of the bombs hit the target (0 hits) can be calculated using the binomial formula:
\[ P(0 \text{ hits}) = \binom{6}{0} p^0 q^6 = 1 \cdot \left(\frac{1}{5}\right)^0 \cdot \left(\frac{4}{5}\right)^6 = \left(\frac{4}{5}\right)^6 \]
- Calculate \( \left(\frac{4}{5}\right)^6 = \frac{4096}{15625} \).

5. **Calculate \( P(1 \text{ hit}) \):**
- The probability that exactly one bomb hits the target:
\[ P(1 \text{ hit}) = \binom{6}{1} p^1 q^5 = 6 \cdot \left(\frac{1}{5}\right)^1 \cdot \left(\frac{4}{5}\right)^5 \]
- Calculate \( 6 \cdot \frac{1}{5} \cdot \left(\frac{4}{5}\right)^5 = 6 \cdot \frac{1}{5} \cdot \frac{1024}{3125} = \frac{6144}{15625} \).

6. **Combine the Probabilities:**
- Now substitute back into the complement formula:
\[ P(\text{at least 2 hits}) = 1 - P(0 \text{ hits}) - P(1 \text{ hit}) = 1 - \frac{4096}{15625} - \frac{6144}{15625} \]
- Combine the fractions:
\[ P(\text{at least 2 hits}) = 1 - \frac{10240}{15625} = \frac{15625 - 10240}{15625} = \frac{5385}{15625} \]

7. **Final Probability:**
- The probability that at least 2 bombs strike the target is:
\[ P(\text{at least 2 hits}) = \frac{5385}{15625} \approx 0.345 \]

### Summary:
The probability that at least 2 bombs strike the target when 6 bombs are dropped is approximately \( 0.345 \).