The probability that a bulb produced in a factory will fuse after 150 days of use in 0.05.
Find the probability that out of 5 such bulbs:
(ii) Not more than one will after 150 days of use.

Answer

To solve the problem, we need to find the probability that out of 5 bulbs, not more than one will fuse after 150 days of use, given that the probability of a bulb fusing is 0.05.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let \( n = 5 \) (the number of bulbs).
- Let \( p = 0.05 \) (the probability that a bulb will fuse).
- Let \( q = 1 - p = 0.95 \) (the probability that a bulb will not fuse).

2. **Define the Random Variable**:
- Let \( X \) be the random variable representing the number of bulbs that fuse after 150 days of use. We want to find \( P(X \leq 1) \).

3. **Calculate \( P(X = 0) \)**:
- The probability that none of the bulbs fuse can be calculated using the binomial probability formula:
\[ P(X = 0) = \binom{n}{0} p^0 q^n = \binom{5}{0} (0.05)^0 (0.95)^5 \]
- Simplifying this gives:
\[ P(X = 0) = 1 \cdot 1 \cdot (0.95)^5 = (0.95)^5 \]

4. **Calculate \( P(X = 1) \)**:
- The probability that exactly one bulb fuses is given by:
\[ P(X = 1) = \binom{n}{1} p^1 q^{n-1} = \binom{5}{1} (0.05)^1 (0.95)^4 \]
- Simplifying this gives:
\[ P(X = 1) = 5 \cdot 0.05 \cdot (0.95)^4 \]

5. **Combine the Probabilities**:
- Now, we can find \( P(X \leq 1) \):
\[ P(X \leq 1) = P(X = 0) + P(X = 1) = (0.95)^5 + 5 \cdot 0.05 \cdot (0.95)^4 \]

6. **Calculate the Values**:
- First, calculate \( (0.95)^5 \):
\[ (0.95)^5 \approx 0.77378 \]
- Next, calculate \( 5 \cdot 0.05 \cdot (0.95)^4 \):
\[ (0.95)^4 \approx 0.81451 \quad \Rightarrow \quad 5 \cdot 0.05 \cdot 0.81451 \approx 0.20363 \]

7. **Final Calculation**:
- Now add the two probabilities:
\[ P(X \leq 1) \approx 0.77378 + 0.20363 \approx 0.97741 \]

### Final Answer:
The probability that out of 5 bulbs, not more than one will fuse after 150 days of use is approximately **0.97741**.