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{w : w has at least two a’s and an odd number of b’s}

I'm having trouble wrapping my head around how to get every single possible outcome based on this language. An odd number of b's is more or less ((bb)b), at least 2 a's is just aa(a*). I know this is not strictly relevant to the final regular expression, it's just how I'm thinking about the language.

Basically I can't figure out the expression for this language. If you need at least 2 a's, then I imagine you'd need 2 a's with no attached kleene star.

Additionally, how do you guarantee an odd number of b's without resorting to something like ((bb)b)?

Actually I can't even figure out where to start with this problem. Writing this question has confused me further.

EDIT:

I've thought of a possible solution:

(a+(bb)*b)*a(a+(bb)*b)*a(a+(bb)*b)*

but this is perhaps a little inelegant

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    Think about $(bb)^*ba(bb)^*ba(bb)^*b$. – 2017-02-07
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    Yeah I just realized I could've done `(a+(bb)*b)*a(a+(bb)*b)*a(a+(bb)*b)*`. I think it looks right, just not sure. – 2017-02-07
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    your solution contains the word w=aa which isn't in the language. – 2017-02-07
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    Maybe not elegant but easy to reason about: you could try to split into cases. you know that every w has two a's, take the first and the last one. now there are 3 places to fill (before the first a, in between, and after the last), think about all the possibilities that create a word with odd number of b's. – 2017-02-07
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    it should lead to the following ugly solution: (bb)*a(a+(bb)*)*a(bb)*b + (bb)*a(a+(bb)*b)*a(bb)* + (bb)*ba(a+(bb)*)*a(bb)* + (bb)*ba(a+(bb)*b)*a(bb)*b – 2017-02-07

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