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The student samples courses without replacement, so I believe this follows the negative hypergeometric distribution. So, $E(X) = \frac{b}{g+1}$, which is smaller than the geometric distribution's expected number of bad courses.
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The student samples courses without replacement, so I believe this follows the negative hypergeometric distribution. So,$E(X) = \frac{b}{g+1}$ , which is smaller than the geometric distribution's expected number of bad courses.
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