Question 225723
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The probability of exactly *[tex \Large k] successes in *[tex \Large n] trials is:


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


where *[tex \Large p] is the probability of success on one trial, *[tex \Large q] the probability of failure on one trial and *[tex \Large \left(n\cr k\right)\ =\ \frac{n!}{k!(n-k)!}]


For your problem you need to know the probability that an odd number occurs zero times or one time or two times.  So you will need the following sum:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_9(k<3)\ =\ \left(P_9(0)\right)\ +\ \left(P_9(1)\right)\ +\ \left(P_9(2)\right)]


Since *[tex \Large p\ =\ q\ =\ \frac{1}{2}], *[tex \Large p^kq^{n-k}\ =\ \left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^6\ =\ \left(\frac{1}{2}\right)^9\ =\ \frac{1}{512}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(9\cr0\right)\ =\ \frac{9!}{0!(9-0)!}\ =\ 1]


(Remember 0! = 1)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(9\cr1\right)\ =\ \frac{9!}{1!(9-1)!}\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(9\cr2\right)\ =\ \frac{9!}{2!(9-2)!}\ =\ \frac{9!}{2!(7)!}\ =\ \frac{9\cdot8}{2}\ =\ 36]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_9(k<3)\ =\ \left(\frac{1}{512}\right)\ +\ \left(\frac{9}{512\right)\ +\ \left(\frac{36}{512}\right)\ =\ \frac{46}{512}\ =\ \frac{23}{256}]


Which is to say that there is better than a 90% chance that it WON'T happen.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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