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:
(iii) More than one will fuse after 150 days of use.

Answer

To solve the problem, we need to find the probability that more than one bulb will fuse after 150 days of use, given that the probability of a bulb fusing is \( p = 0.05 \) and we are considering 5 bulbs.

### Step-by-Step Solution:

1. **Define the Problem**:
Let \( X \) be the random variable representing the number of bulbs that fuse after 150 days. We know that \( X \) follows a binomial distribution with parameters \( n = 5 \) (the number of bulbs) and \( p = 0.05 \) (the probability of a bulb fusing).

2. **Calculate the Probability of No Bulbs Fusing**:
The probability that none of the bulbs fuse (i.e., \( X = 0 \)) can be calculated using the binomial probability formula:
\[ P(X = 0) = \binom{n}{0} p^0 (1-p)^{n} = 1 \cdot (0.05)^0 \cdot (0.95)^5 = (0.95)^5 \]
Now, calculate \( (0.95)^5 \):
\[ (0.95)^5 \approx 0.7738 \]

3. **Calculate the Probability of Exactly One Bulb Fusing**:
Next, we calculate the probability that exactly one bulb fuses (i.e., \( X = 1 \)):
\[ P(X = 1) = \binom{5}{1} p^1 (1-p)^{n-1} = 5 \cdot (0.05)^1 \cdot (0.95)^4 \]
Now, calculate \( 5 \cdot (0.05) \cdot (0.95)^4 \):
\[ (0.95)^4 \approx 0.8145 \]
Thus,
\[ P(X = 1) = 5 \cdot 0.05 \cdot 0.8145 \approx 0.2036 \]

4. **Calculate the Probability of Not More Than One Bulb Fusing**:
Now, we find the probability that not more than one bulb fuses:
\[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.7738 + 0.2036 \approx 0.9774 \]

5. **Calculate the Probability of More Than One Bulb Fusing**:
Finally, the probability that more than one bulb fuses is given by:
\[ P(X > 1) = 1 - P(X \leq 1) \approx 1 - 0.9774 \approx 0.0226 \]

### Final Answer:
The probability that more than one bulb will fuse after 150 days of use is approximately \( 0.0226 \).