Question

The probability that a bulb produced in a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs:

1. None will fuse after 150 days of use.
2. Not more than one will fuse after 150 days of use.
3. More than one will fuse after 150 days of use.
4. At least one will fuse after 150 days of use.

Answer 1

Let:

Probability that a bulb fuses after 150 days = p = 0.05 

Probability that a bulb does not fuse = g = 1 – p = 0.95 

Number of bulbs = n = 5

We use the binomial probability formula:

`P(k) = ((n),(k)) p^kq^(n - k)`

Where k is the number of bulbs that fuse.

i. None will fuse after 150 days

This means k = 0:

`P(0) = ((5),(0)) (0.05)^0 (0.95)^5`

= 1 · 1 · (0.95)⁵ ≈ 0.7738

ii. Not more than one will fuse

 This means k = 0 or k = 1:

`P(0) + P(1) = (0.95)^5 + ((5),(1)) (0.05)^1 (0.95)^4`

= 0.7738 + 5 · 0.05 · (0.95)⁴

= 0.7738 + 5 · 0.05 · 0.8154

= 0.7738 + 0.2038

= 0.9776

iii. More than one will fuse

P(k > 1) = 1 – P(0) – P(1)

= 1 – 0.9776

= 0.0224

iv. At least one will fuse

P(k ≥ 1) = 1 – P(0)

= 1 – 0.7738

= 0.2262