Answer :
Answer:
A. 0.0702
Step-by-step 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:
[tex]P(X \geq 3) = 1- (P(X=0)+P(X=1)+P(X=2))\\[/tex]
For X= 0:
[tex]P(X=0) = (1-0.10)^{10}=0.34868[/tex]
For X= 1:
[tex]P(X=1) = 10*0.1(1-0.10)^{9}=0.38742[/tex]
For X=2
[tex]P(X=2) = \frac{10!}{(10-2)!2!}*0.1^2(1-0.10)^{8}=0.19371[/tex]
Therefore:
[tex]P(X \geq 3) = 1- (0.34868+0.38732+0.19371)\\P(X \geq 3) =0.0702[/tex]
The probability that 3 or more of the calculators will be defective is 0.0702.