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My problem is connected with evaluating $$ P(X\geq n/2)$$ where $X\mathtt {\sim} B(n,p)$. It would be clear for me if in the place of $n/2$ we had $n$. I would appreciate any help.

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    $\begingroup$ Let me think you are trying to get $P(X\geq \frac{n}{2})$ which is $$1-P(X< \frac{n}{2})=\sum_{x=0}^{\frac{n}{2}-1}b(x;n,p)=\sum_{x=0}^{\frac{n}{2}-1}{}^nC_xp^x(1-p)^{n-x}$$ What's wrong with this$?$ $\endgroup$ – emonHR Nov 12 '19 at 19:07
  • $\begingroup$ I have also thought in this way but I wasn't sure. Thanks a lot! $\endgroup$ – Lover Nov 12 '19 at 19:24
  • $\begingroup$ But where is 'one' after first equality? $\endgroup$ – Lover Nov 12 '19 at 20:47
  • $\begingroup$ It's a typo. Ok I will make it answer then $\endgroup$ – emonHR Nov 13 '19 at 6:37
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Let me think you are trying to get $P(X≥\frac{n}{2})$ which is $$P(X≥\frac{n}{2})=1-P(X<\frac{n}{2})\\ =1-\sum_{x=0}^{\frac{n}{2}-1}b(x;n,p)\\=1-\sum_{x=0}^{\frac{n}{2}-1}{}^nC_xp^x(1-p)^{n-x}$$

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Obviously, the answer depends on $n$ and $p,$ so you need to give a formula in $n$ and $p$ as @emonhossain (+1) has done.

If $n = 2$ through $20$ and $p = 1/2,$ then here are two brief tables of results from R. Notice that, for even $n,$ all $P(X \ge n/2)$ are $1/2.$ For large odd numbers, answers become increasingly near to $1/2.$

Notes: (a) Disregard line numbers in brackets. (b) In R, pbinom is a binomial CDF. (c) The small number eps is necessary to get $P(X \ge n/2)$ instead of $P(X > n/2),$ when $n$ is even.

n=2:20; eps = .00001
pr = 1 - pbinom(n/2+eps ,n, .5)
cbind(n, pr)
       n        pr
 [1,]  2 0.2500000
 [2,]  3 0.5000000
 [3,]  4 0.3125000
 [4,]  5 0.5000000
 [5,]  6 0.3437500
 [6,]  7 0.5000000
 [7,]  8 0.3632813
 [8,]  9 0.5000000
 [9,] 10 0.3769531
[10,] 11 0.5000000
[11,] 12 0.3872070
[12,] 13 0.5000000
[13,] 14 0.3952637
[14,] 15 0.5000000
[15,] 16 0.4018097
[16,] 17 0.5000000
[17,] 18 0.4072647
[18,] 19 0.5000000
[19,] 20 0.4119015

A few larger values of $n:$

n=1000:1006;  pr = 1 - pbinom(n/2+eps, n, .5)
cbind(n, pr)

        n        pr
[1,] 1000 0.4873875
[2,] 1001 0.5000000
[3,] 1002 0.4874001
[4,] 1003 0.5000000
[5,] 1004 0.4874126
[6,] 1005 0.5000000
[7,] 1006 0.4874251

By contrast, if $p = 1/3,$ then with increasing $n$ probability tends to lie increasingly below $n/2.$ Thus probabilities $P(X > n/2)$ shrink to $0.$

n = 1:50;  pr = 1 - pbinom(n/2+eps, n, 1/3)
plot(n, pr, type="b", pch=20)
 abline(h=0, col="green2")

enter image description here

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