# Thread: binomial distrib-how to calculate P(X>3 given n=15, success=.40)?

1. ## binomial distrib-how to calculate P(X>3 given n=15, success=.40)?

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

2. Originally Posted by beetz
is it the same as calculating P(X>=3)??
thanks!
no. greater than or equal to 3 and greater than 3 are not the same. x > 3 here means x >= 4. it is the same as calculating:

P(4) + P(5) + ... + P(15)

but the easier way would be to calculate 1 - P(x <= 3) = 1 - P(0) - P(1) - P(2)

3. how would i do that with the calculator....

would that be 1-binomcdf(15,.40,2)?

4. Originally Posted by beetz
how would i do that with the calculator....

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

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

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

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

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

so you want:

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

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

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

${15 \choose 0} = 1$

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

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