# Basics and brainlock

1. Oct 30, 2005

### jimmie 88

(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

2. Oct 30, 2005

### HallsofIvy

Staff Emeritus
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
[tex]\frac{100!}{2(98!)}(.2^2)(.8^{98})[/itex]

Do the arithmetic! (Cancel a lot in those binomial coefficients first.)

Last edited: Oct 30, 2005