The probability that both televisons work is 18/33.
The probability that at least one does not work is 15/33.
Let x = number of shipments, x = 12
P(defective) = 3/12
P(working) = 9/12
The probability that two randomly selected televisions are both working is:
9/12 * 8/11 = 72 / 132 or 18/33
P(2 working) = 18/33
8/11 is used for the second selection since the number of televisions has been reduced due to the first selection.
The probability of 1 not working among the two can be solved through"
P(at least one defective among the 2) = 1 -
P(2 working)
P(at least one defective among the 2) = 15/ 33
Thank you for posting your question. I hope you found you were after. Please feel free to ask me another.
