SOLUTION: Could you please help me with this problem? I do not understand how to find the probability: A retailer received 6 VCR's from the manufacturer of which 3 were damaged during shi

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Question 290904: Could you please help me with this problem? I do not understand how to find the probability:
A retailer received 6 VCR's from the manufacturer of which 3 were damaged during shipment. Two VCRS were sold. What is the probability that one of the customers recieved a damaged VCR?
Many thanks!!
Answer by dabanfield(803) (Show Source):
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Could you please help me with this problem? I do not understand how to find the probability:
A retailer received 6 VCR's from the manufacturer of which 3 were damaged during shipment. Two VCRS were sold. What is the probability that one of the customers recieved a damaged VCR?
Many thanks!!
The probability that exactly one customer (say customer A) gets a bad VCR and the other (say customer B) gets a good one is the sum of the probabilities that customer A gets a good one and customer B gets a bad one OR A gets the bad one and B the good one;
The probablity that A gets a good VCR is 3/6 = 1/2. The probability that customer B gets a bad one given that A got a good one is 3/(6-1) = 3/5. The probability then of this first scenario is (1/2)*(3/5) = 3/10.
The probability for the reverse situation where A gets the bad one and B the good one is the same, 3/10.
So the probability that one or the other of these two scenarios occurs is 3/10 + 3/10 = 3/5.