# A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. - Mathematics and Statistics

Sum

A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector of the retailer picks 20 items from a shipment. What is the probability that the store will receive at most one defective item?

#### Solution

Let X = number of a defective electronic device.

p = probability that a device is defective

∴ p = 3% = 3/100

∴ q = 1 - p = 1 - 3/100 = 97/100

Given: n = 20

∴ X ~ B (20, 3/100)

The p.m.f. of X is given as :

P[X = x] = "^nC_x  p^x q^(n - x)

i.e. p(x) = "^20C_x  (3/100)^x (97/100)^(20 - x)

P (store will receive at most one defective item)

= P[X ≤ 1] = p[X = 0] + P[X = 1]

= p(0) + p(1)

= ""^20C_0 (3/100)^0 (97/100)^(20 - 0) + "^20C_1 (3/100)^1 (97/100)^(20 - 1)

= 1 xx 1 xx (0.97)^20 + 20xx (0.03) xx (0.97)^19

= (0.97 + 0.6)(0.97)^19

= (1.57)(0.97)^19

Hence, the probability that the store will receive at most one defective item = (1.57)(0.97)^19

[Note: Answer in the textbook is incorrect.]

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 8 | Page 254