Question 465798
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You are approaching the problem correctly, but I'm not sure you have the formula appropriately used.


The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


but you want *[tex \Large P_{20}(\geq2,0.5)] which would require calculating:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(\geq2,0.5)\ =\ \sum_{i=2}^{20}\ \left(20\cr\ i\right\)\left(0.5\right)^i\left(1\,-\,0.5\right)^{20\,-\,i}]


A daunting task indeed. However *[tex \Large 1\ -\ P_{20}(\leq1,0.5)] which can be rendered:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_{20}(\leq1,0.5)\ =\ 1\ -\ \left\[\left(20\cr\ 0\right\)\left(0.5\right)^0\left(0.5\right)^{20}\ +\ \left(20\cr\ 1\right\)\left(0.5\right)^1\left(0.5\right)^{19}\right\]\ =\ 1\ -\ 21\,\cdot\,\left(0.5\right)^{20}\ >\ 0.999]


Is much easier to calculate.  In fact, looking a little deeper, you managed to come up with the correct denominator, but your numerator is off by a touch -- should have been 1048576 minus 21 = 1048555.


You only get one problem per posting.  But I'll give you a hint for the second one.  Since you can only move down and right, if any of your first three moves are down, you will miss point B.  Therefore your first three moves must be to the right.  From there you have a limited number of ways to go, so it should be relatively easy to count the number of successful paths.


By the way, I know just by looking at your answer of 60 that it is incorrect without looking at any other part of the problem.  Why do you think that is?  Hint:  You are trying to find a probability.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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