SOLUTION: What is the probability of an 80% free throw shooter making 96 of 100 consecutive free throws?

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Let p=probability of making and q=prob. of missing. p=.8 and q=.2
I am assuming that he must shoot 96 of 100 exactly, no more or less.
We are looking for the 96th term of (p+q)^100 which equals
100C96*p^96q^4 The combination of 100 things taken 96 at a time gives us an easy way to get the large front number.
=3921225*.8^96*.2^4
=.0000031202 or 3.1202*10-6
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Ed