One more

Thanks to you

Attachment 1644

Attachment 1645

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- January 29th 2007, 05:40 PMmauro21plProbability
One more

Thanks to you

Attachment 1644

Attachment 1645 - January 30th 2007, 05:49 AMCaptainBlack
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