Question 1192024
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(a) At the start, there are 10 even-numbered seats out of a total of 20.<br>
P(1st seat # is even) = 10/20
P(2nd seat # is even) = 9/19
...
P(5th seat # is even) = 6/16
P(6th seat # is even) = 5/15<br>
You want all those things to happen, so multiply all those probabilities.  I leave it to you to do the calculations and express the result in the required format.<br>
(b) In this case, 2 of the 10 even-numbered seats are already occupied.  The solution process is the same as in part (a); but now the probabilities are 8/18, 7/17, ..., 4/14, and 3/13.<br>