Katie,
I'll outline how to do problem 1. the other two are very similar.
19 Frozen dinners
10 Pasta
6 Chicken
3 Seafood
Select 5 at random find p(that at least 2 are pasta}
we have 10 pasta and 9 not pasta
First calculate the number of combinations of 2 pastas 10C2 =45 no. of non pasta 9C3 =84
So the total number of combinations having 2 pastas is 45*84=3780
the total number having 3 pastas is 120*36=4320
doing the same calculation for 4 and 5 pastas and adding them up gives 10,242 combinations having 2,3,4 or5 pastas
The total number of combinations of 5 out of 19 is 19C5=11,628
So p(of at least 2 pastas)=10,242/11,628=.881.
Could have calculated the probability of 0 and 1 pastas and subtracted it form one...fewer calculations
Jim