An item that is in 2% of the bills must appear in 2 × 10 ⁸ bills. Across all bills, there are at most (10 ¹⁰ ) × 10 = 10 ¹¹ items mentioned. So at most (10 ¹¹ )/(2 × 10 ⁸ ) = 500 items can appear in 2% of the bills. The number of items in stock is irrelevant.

Out of 100 mails, 10 are spam. The filter will label 9 of 10 spam as spam and 9 of 90 non-spam as spam. So 18 are labelled spam, of which 9 are actually spam. You can compute the same result more formally using conditional probabilities.

Let n be the number of pages visited by the search engine at the time a query is submitted. Assume that it takes constant time to compute the relevance score for each page w.r.t. a query. Then it takes O(n) time to compute the relevance scores, a further O(n) time to build a heap of n relevance scores, and O(k · log n) time for k delete-max operations to return the top k scores.

Let Σ = {x, y, z}. The language in question is (Σ ∗ \ (L ₁ ∪ L ₂ )) ∩ L ₃ where
L ₁ = {w ∈ Σ ∗ | the number of y’s is divisible by 2},
L ₂ = {w ∈ Σ ∗ | the number of y’s is divisible by 7}, and
L ₃ = {w ∈ Σ ∗ | no x appears after a z}.
It suffices to show that all these three languages are regular, and appeal to the fact that regular languages are closed under boolean operations.
They are described by the following regular expressions, respectively.
r ₁ = ((x + z) ∗ y(x + z) ∗ y(x + z) ∗ ) ∗
r ₂ = ((x + z) ∗ y(x + z) ∗ y(x + z) ∗ y(x + z) ∗ y(x + z) ∗ y(x + z) ∗ y(x + z) ∗ y(x + z) ∗ ) ∗
r ₃ = (x + y) ∗ (y + z) ∗