Question 319371
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"between 3 and 6" is ambiguous.  Do you mean inclusive or exclusive of 3 and 6?  No matter, I'll show you what to do either way.



If you mean <i>inclusive</i> of 3 and 6, then the probability you seek is the probability of getting <i>exactly</i> 3 heads out of 10 tosses, plus the probability of getting exactly 4, plus exactly 5, plus exactly 6.


On the other hand, if you mean <i>exclusive</i>, then the probability you seek is the probability of exactly 4 heads out of 10 tosses plus the probability of exactly 5 heads out of 10 tosses.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \left(n\cr k\right\)p^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 has the value *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


For a fair coin toss, the probability of a head on any given trial is *[tex \Large \frac{1}{2}]


So, for the inclusive case, you need to calculate the following:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(3)\ =\ \left(10\cr \,3\right\)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{7}\ =\ \left(10\cr \,3\right\)\left(\frac{1}{2}\right)^{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(4)\ =\ \left(10\cr \,4\right\)\left(\frac{1}{2}\right)^{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(5)\ =\ \left(10\cr \,5\right\)\left(\frac{1}{2}\right)^{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(6)\ =\ \left(10\cr \,6\right\)\left(\frac{1}{2}\right)^{10}]


And sum the four results.


For the exclusive case, you only need to compute


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(4)\ =\ \left(10\cr \,4\right\)\left(\frac{1}{2}\right)^{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(5)\ =\ \left(10\cr \,5\right\)\left(\frac{1}{2}\right)^{10}]


and sum the two results




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