# 20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least o

20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?

I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:

1) the first two are yellow
2) the first two are blue

I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
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Giancarlo Shah
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability 1. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.