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Consider the following statement from Computer Networking A Top-Down Approach textbook,

With packet switching, the probability that a user is active is $0.1$. If there are $35$ users, the probability that there are $11$ or more simultaneously active users is approximately $0.0004$.

$P(11$ or more$) = P(11) + P(12) + \dots + P(35)\\ =1-(P(1) + P(2) + \dots + P(10)) = 1.1111111111111119\times 10^{-11$}$

Something is wrong with my thinking, can someone point it out?

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4 Answers 4

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Let random variable $X$ be the number of actives. Then if we use your procedure, we want $$1-\left(\Pr(X=0)+\Pr(X=1)+\cdots+\Pr(X=10)\right).\tag{1}$$ Note this is a little different from yours, you had left out the possibility $X=0$.

That correction, however, does not account for the error, so you must have computed the probabilities incorrectly. Note that $$\Pr(X=k)=\binom{35}{k}(0.1)^k (0.9)^{35-k}.$$ We have to be very very careful in the computation. The sum we are computing in (1) is very close to $1$, so roundoff error can be catastrophic.

Another way: I can see why you wanted to use (1), it looks like a shorter computation. However, in this situation, it is better to compute $$\Pr(X=11)+\Pr(X=12)+\cdots+\Pr(X=35)\tag{2}$$ directly.

That's because the mean number of actives is $3.5$, so the first number in the sum (2) is small, and pretty quickly we are adding virtually nothing, so we can truncate the computation early. Also, the computation via (2) is numerically stable, while using (1) definitely is not.

Remarks: $1.$ Nowadays, many programs, and even some calculators, will compute binomial probabilities and their sums. Wolfram Alpha (free) will do it.

$2.$ If you want to use an approximate procedure, I would recommend the Poisson approximation. So the required probability is not far from $$\sum_{11}^{35} e^{-3.5} \frac{(3.5)^n}{n!}.$$ Only the first few terms really matter.

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It might be how you computed $P(n)$. It should be $$P(n)= \binom{35}{n} 0.1^n 0.9^{35-n}.$$

Here is the computation in WolframAlpha.

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  • $\begingroup$ right, forgot about bernoulli trials. $\endgroup$
    – user99297
    Sep 4, 2014 at 3:27
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Your random variable follows the binomial distribution $\mathcal{B}(35,0.1)$.

The probability you are asking for is:

$P(X\ge 11)=\displaystyle\sum_{n=11}^{35}\binom{35}{n}\cdot0.1^n\cdot (1-0.1)^{35-n}$.

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The number of users $X$ that are active is Binomial(35, 0.1). So, P(k) = (35 choose k) p^k*(1-p)^(35-k). The summation that you have is hard to compute because you have to be careful in computing (n choose k). A better way is to use a normal approximation to Binomial random variable. We know $\mu = E[X] = 3.5$ and $\sigma^2 = Var[X] = 35*0.1*0.9$. \begin{split} P(X \geq 11) &= P\left( (X-\mu)/\sigma \geq (11 - 3.5)/\sigma \right) \\ &= \Phi(-(11 - 3.5)/\sigma), \end{split} where $\Phi$ is CDF of normal distribution. I got this answer to be 1.191e-05

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