# Conditional Probability

by mtc1973
Tags: conditional, probability
 Sci Advisor HW Helper P: 4,300 For probability 1, it's fairly easy to see that you need at least 6 X. After all, the worst case scenario is that you bind all the inactive sites first, and then the next 3 X that you place are active. For the rest, this is precisely the model that I (and probably a lot of other people too) use when thinking about hypergeometric probabilities. Suppose that you place n X on arbitrary sites. You can do this in $\binom{9}{n}$ ways (the number of ways you can choose n sites to bind to from the 9 available. The probability that k are bound to any of the 3 inactive sites and n - k to active sites is then $$\frac{\binom{3}{k} \binom{6}{n - k}}{\binom{9}{n}}$$