Whenever you see the words "at least one" 
always think of this principle:

P(at least one) = 1 - P(none at all)

So therefore

P(she will open at least 1 can of soup) =

1 - P(she will open no cans of soup) =

1 - P(she will open 2 of the 3 non-soup cans) =

      no. of ways to choose 2 of the 3 non-soup cans
1 - --------------------------------------------------
      no. of ways to choose any 2 of the 5 cans


     C(3,2)        3      7
1 - ------- = 1 - ---- = ---- = 0.7
     C(5,2)        10     10

It's not too hard to check this:

Suppose the 2 cans of vegetables are called V1 and V2 
Suppose the 2 cans of soup are called S1 and S2
Suppose the 1 can of peaches is called P.

So she can choose any of these ten pairs of cans:

 1. {V1,V2}  
 2. {V1,S1}  at least one soup can 
 3. {V1,S2}  at least one soup can  
 4. {V1,P}
 5. {V2,S1}  at least one soup can
 6. {V2,S2}  at least one soup can 
 7. {V2,P}
 8. {S1,S2}  at least one soup can
 9. {S1,P}  at least one soup can
10. {S2,P}  at least one soup can

She will have at least one can of soup if
she chooses 2, 3, 5, 6, 7, 9, or 10. That's
7 out of 10, or a probability of 7/10.

Edwin