Solve Basics & Brainlock (100) (.2)x(.8) 100-x

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The discussion focuses on solving a binomial distribution problem represented by the formula (100 choose x) multiplied by probabilities (.2^x)(.8^(100-x)). Participants clarify that the expression involves binomial coefficients, specifically \frac{100!}{x!(100-x)!}. Examples are provided for x=1 and x=2, demonstrating how to simplify the calculations by canceling terms in the binomial coefficients. The importance of using a calculator for factorials, such as .899!, is noted. The conversation emphasizes the arithmetic involved in solving these types of problems.
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(100) (.2)x(.8) 100-x
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Trying to solve this using 1, then 2 for x. The same parentheses are around both the 100 and the x.
Thanks
 
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Do you mean that first term, in which "The same parentheses are around both the 100 and the x" is a fraction: (100/x) or the binomial coefficient?

Oh, yes, of course, that's a "binomial distribution" calculation:
\frac{100!}{x!(100-x)!}.2^x .8^{(100-x)}

With x= 1 that is
\frac{100!}{99!}(.2)(.8^{99})
All except the last term are easy. You may want to use a calculator for .899!

With x= 2 that is
\frac{100!}{2(98!)}(.2^2)(.8^{98})[/itex]<br /> <br /> Do the arithmetic! (Cancel a lot in those binomial coefficients first.)
 
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