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 \(\frac{100!}{x!(100-x)!} \cdot 0.2^x \cdot 0.8^{(100-x)}\). Participants clarify the interpretation of the equation, specifically addressing the calculations for \(x = 1\) and \(x = 2\). The calculations involve simplifying the binomial coefficients and using a calculator for factorials, particularly for non-integer values like \(0.899!\). The conversation emphasizes the importance of correctly applying the binomial distribution in statistical problems.

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(100) (.2)x(.8) 100-x
x
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:
[tex]\frac{100!}{x!(100-x)!}.2^x .8^{(100-x)}[/tex]

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

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