is it the same as calculating P(X>=3)??

thanks!

- Oct 11th 2007, 05:07 PMbeetzbinomial distrib-how to calculate P(X>3 given n=15, success=.40)?
is it the same as calculating P(X>=3)??

thanks! - Oct 11th 2007, 06:48 PMJhevon
- Oct 11th 2007, 08:33 PMbeetz
how would i do that with the calculator....

would that be 1-binomcdf(15,.40,2)? - Oct 11th 2007, 08:55 PMJhevon
it depends on the calculator. but recall that this is actually what you're doing:

the probability of $\displaystyle k$ successes in $\displaystyle n$ trials is given by:

$\displaystyle P(k) = {n \choose k}p^kq^{n-k}$

where $\displaystyle p$ is the probability of success, and $\displaystyle q = 1 - p$ is the probability of failure.

here, p = 0.4, and q = 0.6 and n = 15

so you want:

$\displaystyle 1 - {15 \choose 0}(0.4)^0(0.6)^{15} - {15 \choose 1}(0.4)^1(0.6)^{14} - {15 \choose 2}(0.4)^2(0.6)^{13}$

now, on my calculator, there is a $\displaystyle \boxed{_nC_r}$ button. so to find say, $\displaystyle {15 \choose 2}$ i would enter, "15, $\displaystyle \boxed{_nC_r}$, 2 = "

but these are easy enough to work by hand if you're not sure about your calculator.

$\displaystyle {15 \choose 0} = 1$

$\displaystyle {15 \choose 1} = \frac {15}1 = 15$

$\displaystyle {15 \choose 2} = \frac {15 \cdot 14}{2 \cdot 1} = 15 \cdot 7 = 105$