a) x1 can be 0 or 1, and for any value of x1 x2 can be 0 or 1, and for any
values of x1, x2 x3 can be 0 or 1,...
there are five x's each of which can be 0 or 1 that is a total of 2^5 outcomes in the sample space.
b) working through them systematicaly and discarding the duplicates
they appear to be:
11000
11100
11010
11001
11110
11011
11101
1111
00110
10110
01110
00111
01111
10111
10101
c) As x4 and x5 are determined there are three bits which may be 0 or 1, so
the number of events in A is 2^3
d) AW is the event that 4 and 5 failed and (1 and 2 are both working or
3 and 4 are both working or 1, 3 and 5 are both working).
That leaves two possibilitiea 1 and 2 working 3 may be 0 or 1, and 4 an 5
failed: 11100, 11000.
RonL