Quick Problem: Binomial Probability Distribution

DavidB

New member
Joined
Feb 18, 2009
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Just a very quick problem that's been annoying me (very much a math lightweight) for a few days:

How do you compute the probability of having 2 boys from a family of 10 children? (Assuming, of course, that boys and girls are the only possible offspring, no hermaphrodites!)

Thank you.
 
C(10,2)(.5)2(.5)8\displaystyle C(10,2)(.5)^{2}(.5)^{8}

Or we can do it this way:

10!8!2!(12)10\displaystyle \frac{10!}{8!2!}\cdot\left(\frac{1}{2}\right)^{10}

which is basically the same thing. See why?.
 
Thank you for your effort.

Do I see why? Well....I'm working on it!

Thank you again. Very prompt and (I'm assuming) correct reply. FreeMathHelp.com is on my "nice" list!
 
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