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we throw 100 Points in range [0,1],

What ist 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>=50) = 1 - P(X=49) this is to unlikely?

Where is my error in reasoning?


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Why multinomial? Binomial(n,p) with n=100 and p=0.2... – Did Nov 9 '11 at 22:18
Virtually $0$. We have a binomial distribution, $100$ trials, probability of success $0.2$. Approximate with the normal mean $20$ variance $100(0.2)(0.8)=16$. So standard deviation is $4$, and $50$ is $7.5$ standard deviation units from the mean. Being $7.5$ standard deviation units or more to the right of the mean basically never happens. The approximation by the normal is pretty good, though it will underestimate tail probabilities. – André Nicolas Nov 9 '11 at 22:21
hmm the cumulated binomial_table for n=100 and p=0.2 shows unsightly values.. – corium Nov 9 '11 at 22:30
$P(x >50)=\sum_{k=51}^{100} \binom{100}{k}(0.2)^k(0.8)^{100-k}$. Wolfram Alpha has no trouble with this, gives about $5.18\times 10^{-12}$. Pretty unlikely. – André Nicolas Nov 9 '11 at 22:37
is this not to incredible? – corium Nov 9 '11 at 22:40
up vote 0 down vote accepted

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.

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$\Pr(X\ge 50)= 1 - \Pr(X \le 49)$. This is not the same as $1-\Pr(X=49)$. It says "$\le$", not "$=$". It is equal to $\sum\limits_{u=0}^{49} \Pr(X=u)$.

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