The probability that a bulb produced by a factory will fuse in 100 days of use is 0.05. Find the probability that out of 5 such bulbs, after 100 days of use :
(i) None fuse.
(ii) Not more than one fuses.
(iii) More than one fuses.
(iv) At least one fuses.

Answer

To solve the problem, we will use the binomial distribution formula. The probability of success (a bulb fusing) is given as \( P = 0.05 \) and the probability of failure (a bulb not fusing) is \( Q = 1 - P = 0.95 \). We are considering 5 bulbs, so \( N = 5 \).

The binomial distribution formula is given by:

\[ P(X = r) = \binom{N}{r} P^r Q^{N-r} \]

where \( \binom{N}{r} \) is the binomial coefficient, \( P \) is the probability of success, \( Q \) is the probability of failure, \( N \) is the number of trials, and \( r \) is the number of successes.

### Step-by-Step Solution:

**(i) Probability that none fuse (r = 0):**

Using the formula:

\[ P(X = 0) = \binom{5}{0} (0.05)^0 (0.95)^{5} \]

Calculating:

\[ \binom{5}{0} = 1 \]
\[ (0.05)^0 = 1 \]
\[ (0.95)^{5} = 0.77378 \text{ (approximately)} \]

Thus,

\[ P(X = 0) = 1 \cdot 1 \cdot 0.77378 = 0.77378 \]

**(ii) Probability that not more than one fuses (r = 0 or r = 1):**

We need to calculate \( P(X = 0) + P(X = 1) \).

First, calculate \( P(X = 1) \):

\[ P(X = 1) = \binom{5}{1} (0.05)^1 (0.95)^{4} \]

Calculating:

\[ \binom{5}{1} = 5 \]
\[ (0.05)^1 = 0.05 \]
\[ (0.95)^{4} = 0.81451 \text{ (approximately)} \]

Thus,

\[ P(X = 1) = 5 \cdot 0.05 \cdot 0.81451 = 0.20363 \text{ (approximately)} \]

Now, summing both probabilities:

\[ P(X \leq 1) = P(X = 0) + P(X = 1) = 0.77378 + 0.20363 = 0.97741 \]

**(iii) Probability that more than one fuses:**

This can be calculated as:

\[ P(X > 1) = 1 - P(X \leq 1) \]

Thus,

\[ P(X > 1) = 1 - 0.97741 = 0.02259 \]

**(iv) Probability that at least one fuses:**

This can be calculated as:

\[ P(X \geq 1) = 1 - P(X = 0) \]

Thus,

\[ P(X \geq 1) = 1 - 0.77378 = 0.22622 \]

### Final Answers:

(i) Probability that none fuse: \( 0.77378 \)

(ii) Probability that not more than one fuses: \( 0.97741 \)

(iii) Probability that more than one fuses: \( 0.02259 \)

(iv) Probability that at least one fuses: \( 0.22622 \)