SOLUTION: suppose you just received a shipment of ten televisions. four of the televisions are defective. if two televisions are randomly selected, compute the probability that both televisi

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Question 1154656: suppose you just received a shipment of ten televisions. four of the televisions are defective. if two televisions are randomly selected, compute the probability that both televisions work. what is the probability that at least one of the two televisions don’t work?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
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Number of televisions = 10
defective 4
selection = 2 n=2
p(defective) = 4/10, p(work) = 6/10=3/5
p(both work)=(3/5)^2 =0.36

p(At least one does not work) (1-0.36)= 0.64

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
suppose you just received a shipment of ten televisions. four of the televisions are defective.
(a) if two televisions are randomly selected, compute the probability that both televisions work.
(b) what is the probability that at least one of the two televisions don’t work?
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(a) The probability in case (a) can be computed in two ways.


    First way is  P = %286%2F10%29%2A%285%2F9%29 = %2830%2F90%29 = 1%2F3.


    Explanation.  

    The probability to select 1-st TV proper is 6%2F10;  the probability to select 2-nd TV proper is 5%2F9.




     Second way is  P = C%5B6%5D%5E2%2FC%5B10%5D%5E2 = %28%28%286%2A5%29%2F2%29%29%2F%28%28%2810%2A9%29%2F2%29%29 = 15%2F45 = 1%2F3,  giving the same answer.

     Explanation.  

     The denominator is the number of all pairs of 10 TVs; the numerator is the number of all pairs of proper TVs.




(b)  The probability in this case is the complement to the probability of case (a), i.e.


          P = 1 - 1%2F3 = 2%2F3.