Its better to write the sample space over those 9 terms because we know the probabilities associated with those 9 terms. Each term is true with half probability and the terms are independent. Hence there are 2^9 = 512 elements in the sample space. An example is TTTTTFFFF i.e. terms 1 to 5 are true and 6 to 9 are false.
Now consider propositions 1. Let E1 be the event that proposition 1 is true.
Give an example of an element from E1.
Define I[E1]: S --> R as
I[E1](x) = 1 if x ∈ E1
= 0 if x ∉ E1
Like wise define I[E2], ....., I[E9].
Interpret N = I[E1] + I[E2] + ... + I[E9]
An example of an element of E1 is x1^~x3^x7 where '^' denotes 'and'.
0=<N<=9 and N=n means that n out of 9 propositions are true. Can't think beyond this.. <smiley image="smiley_unhappy.gif"/>
You can write the sample space as true false values for 7 propositions also and you will have 2^7 elements. But as you pointed out it will be harder (not impossible) and very inconvenient to assign the probability function to the events of the sample space. [Events can be assigned zero probability. So the count will remain 2^7 if you write your sample space this way. In the example you said that proposition 1, 2 and 7 cannot be false at the same time. So the event that all 1, 2 and 7 are false is a zero probability event ].
lol we are posting at the same time. AJ if this is correct then u got the answer right in the first go..if sir says it's right u've got to tell me how u did it :D