# 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, - Mathematics and Statistics

Sum

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 develop any faults within the first 3 years of use.

#### Solution

Let X = number of machines who develops a fault.

p = probability that a machine develops a falt 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)^(40 - 40)

= (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 develop the fault within 3 years of use =  (775.44)(0.003)^38.

Concept: Binomial Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 12 | Page 254