# Binomial distribution greater than

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?

• The complement of the event $\boxed{x\geq 1}$ is the event $\boxed{x=0}$, so your required probability is $1-\Bbb P(x=0)=\dots$ – Prasun Biswas 2 days ago

$$P(x\geq 1)=\sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^x (\frac{1}{2})^{6-x} = \sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^6=(\frac{1}{2})^6 (2^6-1) =0.984375$$ (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.)
$$P(X\geq 1)=1-P(X=0)=1-0.5^6=0.984375$$