Confirming the Sum Notation: P(X ≥ k)

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The Sum Of...Σ

So we were given this awkward question to do out of the textbook... and a classmate and I arrived at similar answers... for part of the question, we are supposed to right a sequence into "sum notation"

Could somebody please tell me if my notation is correct?

Say... we have a binomial random variable... X ~ binomial (n, p)

P(X ≥ k) = P(X=k) + P(X = k +1) + ... + P(X = n)
= (nCk)(p^k)(1-p)^n-k + (nCk+1)(p^k+1)(1-p)^n-(k+1)
= "the sum of i = k to the n" (nCi)(p^i)(1-p)^n-1

I know its messy... but "the sum of i = k to the n" (nCi)(p^i)(1-p)^n-1

but I am looking to see if that's right?

Thanks
 
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oh and by the way... that (nCk) or (nCi) means "N choose k" or "N choose i"
 
\Sigma_{i= k}^n _nC_i p^i(1- p)^{n-i}

NOT "i= k to the n", that would be i= kn!

And it is (1-p)n-i, not (1-p)n-1.
 
oh... how do you make your equations look so nice? I believe that its some softwear that you use?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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