Binomial distribution greater than

Question

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, Feb 23 '21 at 06:37

Cherryblossoms · Accepted Answer · 2021-02-24T05:13:58.943

0

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

edited Feb 24 '21 at 05:13

answered Feb 23 '21 at 06:32

Cherryblossoms

917

That is! could you walk me through your steps? I dont understand how you decided (1/2)^6 and where (2^6−1) came from. – JanusP Feb 23 '21 at 20:27
I edited the answer, hope it is more clear now. – Cherryblossoms Feb 24 '21 at 05:14

score 0 · Answer 2 · answered Feb 24 '21 at 06:39

0

$$P(X\geq 1)=1-P(X=0)=1-0.5^6=0.984375$$

answered Feb 24 '21 at 06:39

tommik

32,733
4
15
34

Binomial distribution greater than

2 Answers2