Question 270296
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What you need is the probability of zero boys in 10 births plus the probability of one plus the probability of two plus the probability of 3.


The probability of exactly *[tex \Large k] successes in *[tex \Large n] trials given the probability of success on any given trial is *[tex \Large p] and the probability of failure on any given trial is *[tex \Large q\ =\ 1\ -\ p] is given by:


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


where *[tex \Large \left(n\cr k\right)] is the number of ways to select *[tex \Large k] things from *[tex \Large n] things when order doesn't matter and is equal to *[tex \Large \frac{n!}{k!\left(n\,-\,k\right)!]


Probabilities are: *[tex \Large p\ =\ P_{boy}\ =\ 0.5] because male and female are equally likely, and therefore *[tex \Large q\ =\ P_{\overline{boy}}\ =\ P_{girl}\ =\ 0.5].


And since you want "at most 3 boys" you need the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}\left(\leq 3\right)\ =\ \sum_{k=0}^3\left(10\cr\,k\right)\left(0.5\right)^k\left(0.5\right)^{10-k}]


Then for your first step, probability of exactly 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}\left(3\right)\ =\ \left(10\cr \,3\right)\left(0.5\right)^3\left(0.5\right)^7]


Then for exactly 2 you need:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}\left(2\right)\ =\ \left(10\cr \,2\right)\left(0.5\right)^2\left(0.5\right)^8]


And then repeat for exactly 1 and exactly 0.  Finally add the 4 results.


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