1. ## Set probability

Let the set X={x1, x2, x3}, if P({x1})=.4 and P({x1,x3})= .5 find probability of P(power set of X) =? ....P= probability.

Attempt. Since probability of a set is the membership function m(x1) for instance, then I thought that probability of the power set will the Sum of probability of all of the sets.
That is P(power set X)= .4+.5+ bunch of 0= .9.
But I am really not sure. if anyone can help it wil be so awesome! I am just confused by this. I'll have a fuzzy set theory like problems to do.

2. Is P{x1 or x2 or x3}=1? In other words, is the occurrence of at least one of the three elements mandatory?

3. This is odd, to me. If X is the sample space, it is unusual to assign a probability to the power set. Probabilities are assigned to subsets of X - elements of the power set (not necessarily all of them though) - and not usually to the power set itself. More explicitly, the power set of the sample space isn't usually in the domain of the probability function.

4. Originally Posted by Sambit
Is P{x1 or x2 or x3}=1? In other words, is the occurrence of at least one of the three elements mandatory?
there is no mendatory probbility on x1, x2, or 3.
So no occurence. Is mandatory, but now that U guys r talking I should recall that P(X) must eqal 1so my attemt definetely weak.

To theodds.
I think the Way u tacle the problem, make sense. so the qstion asked was " determine the probability measure on power set Of X, but I thought this stemnt is eqivalent to P({power set of X}).
So reconsidering the question with same conditions, find probability measure on power set of X. I hope this makes more sense now.

5. Originally Posted by karlito03
there is no mendatory probbility on x1, x2, or 3.
So no occurence. Is mandatory, but now that U guys r talking I should recall that P(X) must eqal 1so my attemt definetely weak.
After reading the above, here is my guess.
Because elementary events are disjoint, $P\left( {\left\{ {x_1 ,x_3 } \right\}} \right) = P\left( {\left\{ {x_1 } \right\}} \right) + P\left( {\left\{ {x_3 } \right\}} \right)$.
Thus that implies that $P\left( {\left\{ {x_3 } \right\}} \right) = 0.1$.
Can you find $P\left( {\left\{ {x_2 } \right\}} \right) = ~?$

6. .4,.5,.1