Question 481883
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This depends on what you consider the probability of a male child on any given trial.  If you are just wanting to do an exercise in calculating a binomial distribution, then you can use 50% as the probabiltiy of a boy on any give trial.  I reality, it doesn't actually work out that way.  See:


<a href="http://www.in-gender.com/XYU/Odds/Gender_Odds.aspx
">Gender Odds</a>


And note that the odds of a girl after having had one or two boys, (or vice versa) varies at each instance of a trial.


If you just need the binomial distribution on a 50/50 per trial basis, then you can either calculate the probability of exactly 1 boy, exactly 2 boys, and exactly 3 boys, then sum the three probabilities, or you can calculate the probability of exactly 3 girls (same as exactly 3 boys) and subtract from 1.  That's because the only way you can't have at least 1 boy is to have all three girls.


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)!}]


For your problem, *[tex \Large n\ =\ 3], *[tex \Large k\ =\ 3], and *[tex \Large p\ =\ 0.5]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_3(3,0.5)\ =\ 1\ -\ \left(3\cr 3\right\)\left(0.5\right)^3\left(0.5\right)^{0}]


Note that *[tex \LARGE \left(n\cr n\right\)\ =\ 1\ \forall\ n\ >\ 0\ \in\ \mathbb{Z}]  and *[tex \LARGE a^0\ =\ 1\ \forall\ a\ \in\ \mathbb{R}]


Next time, flip a coin.  You get the same 50/50 results, but each experiment is a good deal less messy and time consuming compared to having babies.


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