Can anyone tell me how I should calculate this problem?

You are on vacation in a foreign country, and you need help. There are 50 people nearby, and you know 25% of the population of this country speaks English.

If you were to call for help:

(a) What is the probability any of them would understand?
(b) The probability that they will all understand?
(c) What's the probability that none will understand what you're saying?

I've got a feeling this falls under the realm of combinatorics but I don't know how to calculate it. A hand anyone?


This problem wasn't given by a professor, it's just a question that came up while chatting with friends and is being asked out of simple curiosity (actually to settle an argument, but...)

Also I'm not a native English speaker, so that's perhaps why the question wasn't clear enough.

For the sake of simplicity, asume the sample group is made of random people, all equally likely to speak the language or not.

My friend Jerome insists that despite what size the groups is, the chances are still 3 to 1 that no one speaks English, but I differ and gave him the example of 100 coin tosses always resulting in tails. I understand the basic of combinatorics but am in no way a maths expert. So put simply is best. You know, you don't understand it if you can't explain it simply!

Thank you all for taking the time to answer!

  • 2
    $\begingroup$ What exactly about the calculation is unclear to you? This problem is of a standard type, so if you don't give more details about what specifically is confusing you, it may be difficult for someone to even know where to begin to help you. It would be possible to regurgitate the calculations giving you the answers to these questions, but if you don't understand the reasoning behind corresponding calculations then they will not be of much use to you. But since you don't explain what you don't understand, it won't be possible to know which parts of the calculations to explain, and which not to. $\endgroup$ – Chill2Macht Jul 2 '17 at 21:50
  • $\begingroup$ If it's a question of standard type, it's a poorly written one. There is no claim made in the wording about the 50 people nearby being a random sample from the population. Without knowing that, we have no way of answering any of the given questions. $\endgroup$ – G Tony Jacobs Jul 2 '17 at 22:12
  • $\begingroup$ In England, for example, the chances are very high. I think most of the population speaks Arabic now. $\endgroup$ – CogitoErgoCogitoSum Jul 2 '17 at 22:32
  • 1
    $\begingroup$ You are asking three related subproblems at once. These are probability questions, and little insight from combinatorics is needed, but formatting the Question in a way that makes clear what you understand about the three subproblems will help Readers to respond incisively. $\endgroup$ – hardmath Jul 2 '17 at 22:40

Each person has a $.75$ probability of not speaking English. If we can assume that this probability applies independently to each person in the crowd (a very questionable assumption), then we would calculate the probability that none of them speaks English by raising $.75$ to the $50$th power. That yields a very tiny probability, close to $0$.

The probability that at least one of them speaks English would be the complement of this, $1-(.75)^{50}$, which is near certainty. The probability that all of them speak English would be $(.25)^{50}$, which is vanishingly small. Of course, if you run into this particular crowd near the university, the chance of them being an English class goes up. :)

The assumption of independence is a long shot though. If $50$ people are in a crowd together, they're probably not selected at random from the country's population. The fact that they are encountered in the same context makes it more likely that their knowledge of English is correlated among each other.

This is why, when attempting to model real-world situations as Bernoulli experiments, it is important to check all of the assumptions carefully. Whether someone in that crowd knows English depends very strongly on what neighborhood you're in, for example.


This is a Bernoulli-experiment.

When $X$ counts the people which do not speak english, you calculate the propability by:

$$P(X=0)=\binom{50}{0}\cdot (\frac{1}{4})^0\cdot (\frac{3}{4})^{50}$$


The probability of all $50$ not understanding is $(1-.25)^{50}$.

  • $\begingroup$ How is "50" equal to "all do not"? $\endgroup$ – DJohnM Jul 2 '17 at 21:55
  • $\begingroup$ Hopefully the update makes more sense $\endgroup$ – Michael Jul 2 '17 at 22:02
  • $\begingroup$ You might want to learn about the feature of posting mathematical expressions on Math.SE using $\LaTeX$ and MathJax. $\endgroup$ – hardmath Jul 2 '17 at 22:28

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