What’s the difference between having four heads in six trials and four consecutive heads in six trials?
Ans: Twelve desperate points.
Somehow, I misread the my exam question and interpreted it as a geometric distribution problem instead of Pascal distribution problem. I can’t believe that I did
P(X=6) = [(1-p)^(3)]*(p) with p=(0.5)^4
because we need four successive successes, instead of
P(X=6) = (5C3)*(p^4)*[(1-p)^2]
But then, even the way I interpreted it into geometric is wrong. Heck, what am I talking about? When you choose your distribution wrongly, there is no way to do it the right way in the wrong way!
To Prof. Inoue, please apply normal distribution into the grading. I know everyone got at least a B+ for the first midterm but the second midterm is a disaster for the whole support. There is no way anyone in the class will get an A- with the current 95-for-A cut off. I dare bet, letting X is the set of discrete grades (X = {A+, A, A-, B+, B, B-, C+, C, C-, D+, D, D-, E, F}) and P(X=x) is a probability with X=x,
P(X>A-) = 0 and thus P(XA-)
Sigh…
