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It sounds like we have 10 total employees here. Is that right?
Anyway, if that is the case, then you're looking for the probability that the first 6 employees tested do not have the disease, which means the same thing as the probability that the 7th (or higher order) employee has the disease, or \(P(X\ge7)\) with "success" probability 0.4
I'm not infallible. My interpretation of the problem was a bit off. I should have said \(P(X=7)\). Sorry for the confusion!
Tone is hard to distinguish in text. Don't worry about it :)
You're welcome
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