The probability that a bulb produced by a factory will fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5 ) that fuse after 200 days of use . Find the probability of
`(i) X = 0, (ii) X le 1, (iii) X gt 1, (iv) X ge 1.`

Answer

To solve the problem, we will use the binomial probability formula, which is given by:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

where:
- \( n \) is the total number of trials (in this case, 5 bulbs),
- \( k \) is the number of successes (bulbs that fuse),
- \( p \) is the probability of success (bulb fuses),
- \( q \) is the probability of failure (bulb does not fuse).

Given:
- \( p = 0.2 \) (probability that a bulb fuses),
- \( q = 1 - p = 0.8 \) (probability that a bulb does not fuse),
- \( n = 5 \) (total number of bulbs).

Now, we will find the probabilities for each part of the question:

### (i) Find \( P(X = 0) \)

Using the binomial formula:

\[ P(X = 0) = \binom{5}{0} (0.2)^0 (0.8)^{5} = 1 \cdot 1 \cdot (0.8)^5 \]

Calculating \( (0.8)^5 \):

\[ (0.8)^5 = 0.32768 \]

Thus,

\[ P(X = 0) = 0.32768 \]

### (ii) Find \( P(X \leq 1) \)

This is the sum of the probabilities for \( X = 0 \) and \( X = 1 \):

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

We already calculated \( P(X = 0) \). Now we calculate \( P(X = 1) \):

\[ P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^{4} = 5 \cdot (0.2) \cdot (0.8)^4 \]

Calculating \( (0.8)^4 \):

\[ (0.8)^4 = 0.4096 \]

Thus,

\[ P(X = 1) = 5 \cdot 0.2 \cdot 0.4096 = 5 \cdot 0.08192 = 0.4096 \]

Now, adding both probabilities:

\[ P(X \leq 1) = 0.32768 + 0.4096 = 0.73728 \]

### (iii) Find \( P(X > 1) \)

This can be calculated as:

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

Thus,

\[ P(X > 1) = 1 - 0.73728 = 0.26272 \]

### (iv) Find \( P(X \geq 1) \)

This can be calculated as:

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

Thus,

\[ P(X \geq 1) = 1 - 0.32768 = 0.67232 \]

### Summary of Results:
- \( P(X = 0) = 0.32768 \)
- \( P(X \leq 1) = 0.73728 \)
- \( P(X > 1) = 0.26272 \)
- \( P(X \geq 1) = 0.67232 \)