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.