SOLUTION: A shipment of pins contains 25 good ones and 2 defective ones. At the receiving
department an inspector picks three pins at random and tests them. If any
defective pin is found
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Question 357303: A shipment of pins contains 25 good ones and 2 defective ones. At the receiving
department an inspector picks three pins at random and tests them. If any
defective pin is found among the three that are tested, the shipment would be
rejected. What is the probability that the shipment would be accepted ? To
increase the probability of acceptance to atleast 90%, it is decided to do one of the
following:
1. Add some good pins to the shipment
2. Remove some defective pins in the shipment
For each of the two options, find out exactly how many pins should be added or
removed.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
(23/25)^3=.7787 probability that the shipment would be accepted.
.
1.
Let x = number of pins to be added to increase the probability of acceptance to atleast 90%.
((23+x)/(25+x))^3>=.9
((23+x)/(25+x))>=.9655... Cube root of each side.
23+x>=.9655(25+x)
23+x>=24.137+.9655x
.0345x=1.137
x>=33 pins should be added.
.
2.
(23/24)^3=.88 one defective pin removed.
Both defective pins would have to be removed to increase the probability of acceptance to atleast 90%.
.
Ed
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