# Binomial problem

#### nacho

##### Active member
please refer to attached image

I don't understand the reasoning for i=6

since a qualified candidate must have answered 15+ questions correctly, we have
i being summed from 15 to 20 for the first part.

shouldn't the second part be summed from i=15 to 20?

for example, $(20,6)$ would imply, 6 successes, and 14 failures in a question. This would make a candidate unsuccessful by the quiz and, would give us $P(A^c|Q^c)$ instead of $P(A|Q^c)$ ?

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