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**ANDS!** 1. Pretty much the same as the other question. The probability that there will be enough seats, is the compliment of the probability that there WONT be enough seats - i.e. 271 or more people show up. This problem can be done one of two ways:

Using the Binomial distribution and calculating 29 binomial equations, where N is equal to 300 tickets and K is equal to going through those 300 tickets and choosing "29, 28, 27. . .0" that are going to no show;

Or you can remember that you can approximate the Binomial using the Normal as in the problem from your other thread (we can use the normal here since 5<np, and nq).

2. Remember:

$\displaystyle V(X)=\Sigma x^{2}P(X=x)-\mu^{2}_x$