Question #106547
Suppose you just received a shipment of
twelve
twelve televisions.
Three
Three of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?
1
Expert's answer
2020-03-26T17:00:25-0400

P(both televisions work) = (92)(122)=36660.54\frac{{9\choose2}}{{12\choose 2}} = \frac{36}{66} \approx 0.54. This is because there are 9 televisions that work out of 12 and we are choosing 2 of them.


P(at least one of the televisions does not work) = 1-P(both televisions work) \approx 1-0.54 = 0.46


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