Quicksort where element comparison outcome is random. Probability of element being in a certain position
In quick sort, after a partition, you have placed every element smaller than your pivot p below p and every element larger than p above it. Therefore after a partition you know the pivot is in the correct position and therefore it will not be moved again. Since you choose the last element (M8) as your pivot, this simplifies the process, since you only have to go through one iteration.

When you compare each element to M8, you will randomly be told that M8 is greater or the other element is "greater" each with 1/2 odds, and if the other element is "greater" it will be placed to the right of M8. Therefore, for M8 to end up in position 0, 1, or 2 it will have to have 2 or less elements that are determined to be "smaller" than M8. We can determine this probability through a geometric sequence:

0: (0.5^7)(0.5^0)(C(7,7)) = (0.5^7)(1) = .0078125

1: (0.5^6)(0.5^1)(C(7,6)) = (0.5^7)(7) = .0546875

2: (0.5^5)(0.5^2)(C(7,5)) = (0.5^7)(21) = .1640625