Question

The probability that a machine develops a fault within the first 3 years of use is 0.003. If 40 machines are selected at random, calculate the probability that 38 or more will not develop any faults within the first 3 years of use.

Answer 1

Let X = number of machines that develop a fault.

p = probability that a machine develops a fault within the first 3 years of use

∴ p = 0.003

and q = 1 - p = 1 - 0.003 = 0.997

Given: n = 40

∴ X ~ B (40, 0.003)

The p.m.f. of X is given by

P(X = x) = `"^nC_x  p^x  q^(n - x)`, x = 0, 1, 2,...,n

i.e. p(x) = `"^40C_x  (0.003)^x  (0.997)^(40 - x)`, x = 0, 1, 2, ....,40

P(38 or more machines will develop any fault)

= P(X ≥ 38) = P(X = 38) + P(X = 39) + P(X = 40)

= p(38) + p(39) + p(40)

`= ""^40C_38 (0.003)^38 (0.997)^(40 - 38) + ""^40C_39 (0.003)^39 (0.997)^(40 - 39) + "^40C_40 (0.003)^40 (0.997)^0`

`= (40 xx 39)/(2 xx 1) (0.003)^38 (0.997)^2 + 40(0.003)^39 (0.997)^1 + 1 * (0.003)^40 (0.997)^0`

`= (780)(0.003)^38 (0.997)^2 + (40) (0.003)^39 (0.997) + 1 xx (0.003)^40 xx 1`

`= (0.003)^38 [(780)(0.997)^2 + 40(0.003)(0.997) + (0.003)^2]`

`= (0.003)^38 [775.327 + 0.1196 + 0.000009]`

`= (0.003)^38 (775.446609)`

`= (775.446609)(0.003)^38`

`≈ (775.44)(0.003)^38`

Hence, the probability that 38 or more machines will not develop the fault within 3 years of use = ` (775.44)(0.003)^38`.