Originally Posted by

**Gretchy** Hmm.

I know that $\displaystyle \binom{n}{x}p^x(1-p)^{n-x}$ where $\displaystyle \binom{n}{x} = \frac{n!}{x!(n-x)!}$, then

$\displaystyle \binom{12}{3} = \frac{12!}{3!(9!)} = 220$

Plug it in and...

$\displaystyle 220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]$

$\displaystyle = 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}] $

$\displaystyle = 220\cdot [(0.6)^{3}\cdot(0.4)^{9}] $

$\displaystyle = 220\cdot [(0.216)\cdot(0.000262)] $

$\displaystyle = 220 \cdot (0.000056592) $

$\displaystyle = 0.01245024$

Similarly,

$\displaystyle \binom{8}{3} = \frac{8!}{3!(5!)} = 56$

$\displaystyle 56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}] $

$\displaystyle = 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]$

$\displaystyle = 56\cdot [(0.064)\cdot(0.7776)] $

$\displaystyle = 56\cdot(0.0497664) $

$\displaystyle = 2.7869184$

So... I add them? $\displaystyle (3 + 3 = 6)$

$\displaystyle 2.7869184 + 0.01245024 = 2.79936864$

Yes? No?