The probability that a bulb produced by a factory will fuse after 160 days of use is 0.06. Find the probability that out of 5 such bulbs at the most one bulb will fuse after 160 days of use.

Answer

To solve the problem, we need to find the probability that at most one bulb will fuse out of five bulbs, given that the probability of a bulb fusing after 160 days is 0.06.

### Step-by-step Solution:

1. **Define the probabilities**:
- Let \( p = 0.06 \) (the probability that a bulb will fuse).
- Let \( q = 1 - p = 1 - 0.06 = 0.94 \) (the probability that a bulb will not fuse).

2. **Identify the scenario**:
- We are looking for the probability that at most one bulb fuses out of five. This can be expressed as:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) \]
- Here, \( X \) is the random variable representing the number of bulbs that fuse.

3. **Use the binomial probability formula**:
- The probability of exactly \( r \) successes (bulbs fusing) in \( n \) trials (bulbs) is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
- In our case, \( n = 5 \), \( p = 0.06 \), and \( q = 0.94 \).

4. **Calculate \( P(X = 0) \)**:
- For \( r = 0 \):
\[ P(X = 0) = \binom{5}{0} (0.06)^0 (0.94)^5 = 1 \cdot 1 \cdot (0.94)^5 \]
- Calculate \( (0.94)^5 \):
\[ (0.94)^5 \approx 0.7351 \]
- Thus, \( P(X = 0) \approx 0.7351 \).

5. **Calculate \( P(X = 1) \)**:
- For \( r = 1 \):
\[ P(X = 1) = \binom{5}{1} (0.06)^1 (0.94)^4 = 5 \cdot (0.06) \cdot (0.94)^4 \]
- Calculate \( (0.94)^4 \):
\[ (0.94)^4 \approx 0.7820 \]
- Thus,
\[ P(X = 1) = 5 \cdot 0.06 \cdot 0.7820 \approx 5 \cdot 0.04692 \approx 0.2346 \]

6. **Combine the probabilities**:
- Now, we can find \( P(X \leq 1) \):
\[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.7351 + 0.2346 \approx 0.9697 \]

### Final Answer:
The probability that at most one bulb will fuse after 160 days of use is approximately \( 0.9697 \).