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:
(i) None will fuse after 150 days of use.

Answer

To solve the problem, we need to find the probability that none of the 5 bulbs 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 Probability of a Bulb Not Fusing**:
- The probability that a bulb will fuse after 150 days is given as \( P(\text{fuse}) = 0.05 \).
- Therefore, the probability that a bulb will not fuse is:
\[ P(\text{not fuse}) = 1 - P(\text{fuse}) = 1 - 0.05 = 0.95 \]

2. **Define the Random Variable**:
- Let \( X \) be the number of bulbs that fuse out of 5 bulbs. We are interested in finding \( P(X = 0) \), the probability that none of the bulbs fuse.

3. **Use the Binomial Probability Formula**:
- The scenario can be modeled using the binomial distribution, where:
- \( n = 5 \) (the number of trials, or bulbs),
- \( p = 0.05 \) (the probability of success, i.e., a bulb fusing),
- \( q = 0.95 \) (the probability of failure, i.e., a bulb not fusing).
- The probability mass function of a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- For our case, we want \( k = 0 \):
\[ P(X = 0) = \binom{5}{0} (0.05)^0 (0.95)^{5} \]

4. **Calculate the Binomial Coefficient**:
- The binomial coefficient \( \binom{5}{0} = 1 \).

5. **Substitute and Simplify**:
- Now substituting the values into the formula:
\[ P(X = 0) = 1 \cdot (0.05)^0 \cdot (0.95)^{5} = 1 \cdot 1 \cdot (0.95)^{5} = (0.95)^{5} \]

6. **Calculate \( (0.95)^{5} \)**:
- Now we compute \( (0.95)^{5} \):
\[ (0.95)^{5} = 0.77378125 \]
- Rounding this, we get approximately \( 0.7738 \).

7. **Final Answer**:
- Therefore, the probability that none of the 5 bulbs will fuse after 150 days of use is approximately:
\[ P(X = 0) \approx 0.7738 \]