Question

The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 10%.If 10 calculators are selected at random, what is the probability that 3 or more of the calculators will be defective?

A. 0.0702
B. 0 .2639
C. 0.0016
D. 0

Answer 1

A. 0.0702

Explanation:

The probability that 3 or more of the calculators will be defective is 1 minus the probability that exactly zero, one or two calculator are defective:

`P(X \geq 3) = 1- (P(X=0)+P(X=1)+P(X=2))\\`

For X= 0:

`P(X=0) = (1-0.10)^(10)=0.34868`

For X= 1:

`P(X=1) = 10*0.1(1-0.10)^(9)=0.38742`

For X=2

`P(X=2) = (10!)/((10-2)!2!)*0.1^2(1-0.10)^(8)=0.19371`

Therefore:

`P(X \geq 3) = 1- (0.34868+0.38732+0.19371)\\P(X \geq 3) =0.0702`

The probability that 3 or more of the calculators will be defective is 0.0702.