The probability of a bit error in a communication line is $10^{-5}$ per bit. Suppose we examine a string of $1000$ independent bits. Calculate the probability of $0$, $1$, $2$, and $3$ errors in the string using Poisson theory and using binomial theory. Compare the results.

  • $\begingroup$ What have you tried ? Why can't you compute $P_0 + P_1 + P_2 + P_3$ in the specified ways ? $\endgroup$ Jun 19 '15 at 11:42
  • $\begingroup$ I think I got it, I did what I explained below in my reply to calculus' answer. If I am supposed to add the probabilities together, however, I've probably done it wrong. $\endgroup$
    – Multiplier
    Jun 19 '15 at 12:12


binomial distribution: $P(X=x)={n \choose x}\cdot p^x\cdot (1-p)^{n-x}$

$n=1000; p=10^{-5}, 1-p=1-10^{-5}$

poisson distribution: $P(X=x)=e^{-\lambda }\cdot \frac{\lambda ^x}{x!}$

$\lambda=n\cdot p=10^{-5}\cdot 1000=\frac{1}{100} $

Both distributions can be approximated by the normal distribution. In the case of a binomial distributed variable: $$P(X=x)=\Phi \left( \frac{x+0.5-n\cdot p}{\sqrt{n\cdot p \cdot (1-p)}} \right)-\Phi \left( \frac{x-0.5-n\cdot p}{\sqrt{n\cdot p \cdot (1-p)}} \right)$$

$\Phi(\cdot )$ is the cummulative function of the standard normal distribution

In the case of a poisson distributed variable: $$P(X=x)=\Phi \left( \frac{x+0.5-\lambda}{\sqrt{\lambda}} \right)-\Phi \left( \frac{x-0.5-\lambda}{{\sqrt{\lambda}}} \right)$$

There are some conditions for approximation. They are only rule of thumbs.

Binomial distribution: $n\cdot p\geq 10$ and $n\cdot (1-p)\geq 10$

Poisson Distribution: $\lambda >10$

  • $\begingroup$ Thank you very much. It makes sense once you know what the values of the parameters are. Then I just replaced $x$ with $0$, $1$, $2$, and $3$ and calculated the values with the pdf formulas of the Poisson and Binomial distributions, getting basically the same answers. I think the factorials in the Binomial distribution became too big for my calculator so I had to use Wolfram Alpha for it. Is that why the Poisson is used to approximate the binomial? Because it is much easier to compute values? $\endgroup$
    – Multiplier
    Jun 19 '15 at 12:11
  • $\begingroup$ Yes, you are right, that´s why you have to approximate the binomial by the poisson distribution. This works only for "small" x. $x!$ is hard to calclulate, if x is "large" $\endgroup$ Jun 19 '15 at 12:26
  • $\begingroup$ Ok, I think I understand. Because only if $p$ is small will the mean of the exponential distribution be large and the $e^{\lambda x}$ small, so that they balance each other out. So is it possible to approximate any distribution with another distribution as long as the curve shapes of the pdf look similar, regardless of whether one is a discrete distribution and one a continuous distribution? $\endgroup$
    – Multiplier
    Jun 19 '15 at 12:37
  • $\begingroup$ The expoenential distributioin was not a good idea. I have made an edit. $\endgroup$ Jun 19 '15 at 13:52
  • $\begingroup$ You don't have to compute $1000!$ to compute $1000 \choose k$. For say $k=3$ it is just $1000*999*998/3!$. The factors in the denominator cancel most of the factors in the numerator. $\endgroup$
    – BruceZ
    Jun 20 '15 at 0:07

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