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

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one

will fuse after 150 days of use.

#### Solution

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05

∴ q = 1 - p = 1- 0.05 = 0.95`

X has a binomial distribution with n = 5 and p = 0.05

∴ P (X = x) = `"^nC_x p^x q^(n - x)` where x = 1, 2, ....n

`= "^5C_x (0.05)^x (0.95)^(n - x)`

(i) **P (none)** = P(X = 0)

= p(0) = `"^5C_0 (0.05)^0 (0.95)^(5 - 0)`

`= 1xx1 xx (0.95)^5`

= (0.95)^{2}

(ii) **P (not more than one)** = P(X ≤ 1)

= p(0) + p(1)

`= ""^5C_0 *(0.05)^0 (0.95)^(5-0) + "^5C_1 (0.05)^1 (0.95)^4`

`= 1xx1 xx (0.95)^5 + 5 xx (0.05) xx (0.95)^4`

`= (0.95)^4 [0.95 + 5(0.05)]`

`= (0.95)^4 (0.95 + 0.25)`

`= (0.95)^4 (1.20) = (1.2)(0.95)^4`

(iii) **P (more than 1)** = P(X > 1)

= 1 - P[X ≤ 1]

= 1 - (1.2)(0.95)^{4}

(iv) **P (at least one)** = P(X ≥ 1)

= 1 - P[X = 0]

= 1 - p(0)

= 1 - `"^5C_0 (0.05)^0 (0.95)^(5 - 0)`

`= 1 - 1 xx 1 xx (0.95)^5`

`= 1 - (0.95)^5`