If $n = 6$ and $p = 0.50$, what is the probability that $x ≥ 1$?
$P(x ≥ 1 | n = 6 \,\text{and}\, p = 0.50) = ?$
Attempt:
$p=0.5$
$n=6$
$x=1 \,\text{or}\, 2$ etc
$(\frac{n!}{x!(n-x)!})(p^x(1-p)^{n-x})$
So I calculated 0.0938 for 1 and 2,3 etc etc but none of the answers were right, what am i missing and need to do to arive at the answer?
$P(x\geq 1)=\sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^x (\frac{1}{2})^{6-x} = \sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^6=(\frac{1}{2})^6 (2^6-1) =0.984375$ (by identity).
(Since, $(1+y)^n=^nC_0+^nC_1 y +^nC_2 y^2+...+^nC_n y^n$, take $y=1$ to get the above identity.)
Let me know if this is correct.