Box with chocolates probability 
I am given a box with 20 chocolates, all of them identical from the outside but 5 of them are with cherry filling, 7 with cream and 8 with nuts.
  I ate 10 at random. What is the probability I ate at least one of each kind?

My attempt: All different ways to select 10 chocolates out of 20 are 
$C(20,10)$.
Then the required probability is 
$1 - ( C(12,10)+C(13,10)+C(15,10) ) / C(20,10)$
I am getting about 98% but this doesn't seem correct to me...
Any ideas?
 A: Let $A$ be the event that I did not eat any with cherry filling.
Let $B$ be the event that I did not eat any with cream.
Let $C$ be the event that I did not eat any with nuts.
Then with inclusion/exclusion we find:
$$P\left(A^{\complement}\cap B^{\complement}\cap C^{\complement}\right)=1-P\left(A\cup B\cup C\right)=$$$$1-P\left(A\right)-P\left(B\right)-P\left(C\right)+P\left(A\cap B\right)+P\left(A\cap C\right)+P\left(B\cap C\right)-P\left(A\cap B\cap C\right)$$
But the events that differ from $A,B,C$ are all empty so that we can
simplify:
$$P\left(A^{\complement}\cap B^{\complement}\cap C^{\complement}\right)=1-P\left(A\right)-P\left(B\right)-P\left(C\right)=\frac{\binom{20}{10}-\binom{15}{10}-\binom{13}{10}-\binom{12}{10}}{\binom{20}{10}}$$
showing that your solution is correct.
A: Yeah, you're right, in general case, you have to use inclusion-exclusion principle to account for situations where you take only one type of sweet for example but no worries here because there is no possibility to choose from only one type.
In general we have something like:
P(not all types taken) = P(taken only from 1 and 2) + P(taken only from 2 and 3) + P(taken only from 3 and 1) - P(taken only from 1) - P(taken only from 2) - P(taken only from 3).
And to feed your intuition, check expected value of each of chocolate type.
