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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?

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  • $\begingroup$ The complement of the event $\boxed{x\geq 1}$ is the event $\boxed{x=0}$, so your required probability is $1-\Bbb P(x=0)=\dots$ $\endgroup$ – Prasun Biswas 2 days ago
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$P(x\geq 1)=\sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^x (\frac{1}{2})^{6-x} = \sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^6=(\frac{1}{2})^6 (2^6-1) =0.984375$ (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.

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  • $\begingroup$ 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. $\endgroup$ – Admiral Pyro 2 days ago
  • $\begingroup$ I edited the answer, hope it is more clear now. $\endgroup$ – user812951 yesterday
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$$P(X\geq 1)=1-P(X=0)=1-0.5^6=0.984375$$

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