SOLUTION: HI, i need help with this problem: A new shipment of 9 VCRs, which will be offered at a very low price, are delivered to a store. Unknow to the manager, 4 of the VCRs are defecti

Algebra ->  Probability-and-statistics -> SOLUTION: HI, i need help with this problem: A new shipment of 9 VCRs, which will be offered at a very low price, are delivered to a store. Unknow to the manager, 4 of the VCRs are defecti      Log On


   

Question 65432: HI, i need help with this problem:
A new shipment of 9 VCRs, which will be offered at a very low price, are delivered to a store. Unknow to the manager, 4 of the VCRs are defective.
Determine the probability that at least one of the first 3 VCRs sold is defective. Express your answer to the nearest tenth of one percent.:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
A new shipment of 9 VCRs, which will be offered at a very low price, are
delivered to a store. Unknow to the manager, 4 of the VCRs are defective.
Determine the probability that at least one of the first 3 VCRs sold is
defective. Express your answer to the nearest tenth of one percent.: 

We use the complement event.

P(choosing a bad VCR) = 4/9

P(choosing a good VCR) = 1 - P(choosing a bad VCR) = 5/9

So

P(choosing at least one defective in the three) = 

1 - P(choosing 3 good VCR's) = 

1 - P(choosing 1st good AND choosing 2nd good AND choosing 3rd good) =

1 - P(choosing 1st good) times
          P(choosing 2nd good given that the 1st one was good) times
                P(choosing 3rd good given that the 1st and 2nd were good) 

= 1 - (5/9)(4/8)(3/7) = 1 - 5/42 = 37/42 = .880952381 = 88% approximately

Edwin