You and two friends are selected for jury duty along with 27 others. Of you 30 tweleve will be selected and eighteen dismissed. Presuming each of the 30 has an equal probability of selection or dismissal determine the following probabilities:
A. You will be selected
B. You will be dismissed
C. All three of you are selected
D. All three of you are dismissed

You can think this through by imaging there are 30 chairs in the room, and whoever sits in any of the first 12 chairs will be on the jury. When you walk in and sit in a random chair you will be on the jury if you happen to sit in any of the first 12 out of 30 chairs, so the probability of you being on the jury is 12/30 = 40%. Conversely, the probability of you not being on the jury is equaklto the probability of sitting in one of the last 18 chairs, or 18/30 = 60%.

For all three friends to be selected the first person to take a chair must be in one of the 12, the second must be in one of the remaining 11 or 29, and the third must be in one of the remaining 10 out of 28. So the probabuility of all 3 being chosen in 12/30 x 11/29 x 10/28. Can you apply this same reasoning to answer part D?