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I have a binomial distribution, with Random Variable Y and n trials. r is an integer. How can I show that P(Y ≥ r) = P(X ≤ n − r), such that Y is a random variable with probability of success p, and X is a random variable with probability of success (1- p)? I have tried expressing P(Y ≥ r) as: $$\sum_{x=r}^{n} \binom{n}{x} p^x(1-p)^{n-x} = \sum_{x=0}^{n-r} \binom{n}{x+r} p^{x+r}(1-p)^{n-(x+r)}$$ I need to come up with this in the end: $$\sum_{x=0}^{n-r} \binom{n}{x} p^{n-x}(1-p)^{x}$$ but I am unsure of how to proceed next. Could someone give me a hint?

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Recall the identity $$\binom nk = \binom n{n-k}. $$ So $$\mathbb P(Y\geqslant r)=\sum_{k=r}^n\binom nk p^k (1-p)^{n-k} = \sum_{k=r}^n\binom n{n-k}p^k(1-p)^{n-k}. $$ Set $j=n-k$, then the above is equal to $$\sum_{j=0}^{n-r} \binom nj (1-p)^j p^{n-j}, $$ which is $\mathbb P(X\leqslant n-r)$.

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  • $\begingroup$ I don't understand how you are changing the bounds of the sum. Could you please explain how you do it so that the expression is still valid? $\endgroup$ – roro172 Feb 6 '15 at 0:08
  • $\begingroup$ When $j=0$, $k=n$. When $j=n-r$, $k=r$. You're summing the same terms, just in reverse order. $\endgroup$ – Math1000 Feb 6 '15 at 0:17

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