Multinomial theorem for $P(X>50)$ We throw $100$ Points in range $[0,1]$.
What is the probability that more than $50$ Points touch down in $[0,0.2)$?
E: "Point right" --> $p= 0.2$
F: "Point fail" ---> $q= 0.8$
Multinomial theorem
$P(X\geq50) = 1 - P(X=49)$
this is to unlikely? 
Where is my error in reasoning?
thx
 A: $\Pr(X\ge 50)= 1 - \Pr(X \le 49)$.  This is not the same as $1-\Pr(X=49)$.  It says "$\le$", not "$=$".  It is equal to $\sum\limits_{u=0}^{49} \Pr(X=u)$.
A: Here on any trial, the probability of picking a point in $[0,0.2)$ is $0.2$. Call picking such a point a "success."  Let the experiment be repeated $100$ times. If the random variable $X$ is the number of successes, then $X$ has binomial distribution. The probability that the number of successes is greater than $50$ is
$$\sum_{k=51}^{100} \binom{100}{k}(0.2)^k(0.8)^{100-k}.$$
In the old days, the above sum would have been very unpleasant to compute. However, Wolfram Alpha computes this easily, and gives an answer of about $5.18\times 10^{-12}$.  
Comment: In the post, there is the assertion that $P(X\ge 50)=1-P(X=49)$. That is not correct. Indeed $P(X \ge 50)$ is close to $0$, while $1-P(X=49)$ is close to $1$. But the following assertion would be correct.
$$P(X \ge 50)=1-P(X\le 49).$$
By the way, Wolfram Alpha says that $P(X \ge 50)\approx 2.14\times 10^{-11}$. Thus $P(X=50)\approx 1.62\times 10^{-11}$. This is very small, of course, but about $3$ times as large as the probability that $X\ge 51$.  That shows that the terms after $50$ are not only small, they also decay rapidly. 
