Question 75236: If 10% of the rivets produced by a machine are defective, find the probability that out of 5 randomly chosen rivets:
(i) one or more will be defective
(ii)at most two will be defective
It would be really helpful for me if you could answer this question. Thank you.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If 10% of the rivets produced by a machine are defective, find the probability that out of 5 randomly chosen rivets:
The problem is "binomial" because the rivets are defective or not defective.
P(defective)=0.10; P(not defective)=0.90
(i) P(one or more will be defective) = 1 - P(none are defective)
1-(0.90)^5
= 0.40951
-----------
If you use a TI calculator it is "1 - binompdf(5,0.1,0)"
========================
(ii)P(at most two will be defective)
=P(x=0)+P(x=1)+P(x=2)
=binomcdf(5,0.1,2)=0.99144
==========
Cheers,
Stan H.
|
|
|
