Probability of Not Getting Candy

[alfred2]

Hi I've found this exercise but I do not understand the solution:
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In a party c candies are given randomly to n kids. ¿Which is the probablility that my nephew doesn't receive any candy? We supose the candies and the kids are numerated. Each of the candies can be given to any of the n kids, so there are n^c possible cases and the unfavorable ones for my nephew are all the manners to distribute the candies between the (n-1) remaining kids i.e. (n-1)^c. So the probability is (1-1/n)^c. If n=c the probability is practically independent of n, being aproximately equals to e^-1=0.37
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First of all I think there's a mistake and the probability is (n-1/n)^c and not (1-1/n)^c. Then what's the meaning of "If n=c the probability is practically independent of n"? What does independent mean in this case? And how do we obtain e^-1? Thanks! If anyone know a book as "probability and statistics for dummies" please let me know! Thanks ;)

 

[I like Serena]

Hi I've found this exercise but I do not understand the solution:
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In a party c candies are given randomly to n kids. ¿Which is the probablility that my nephew doesn't receive any candy? We supose the candies and the kids are numerated. Each of the candies can be given to any of the n kids, so there are n^c possible cases and the unfavorable ones for my nephew are all the manners to distribute the candies between the (n-1) remaining kids i.e. (n-1)^c. So the probability is (1-1/n)^c. If n=c the probability is practically independent of n, being aproximately equals to e^-1=0.37
_________________________________________________________________
First of all I think there's a mistake and the probability is (n-1/n)^c and not (1-1/n)^c. Then what's the meaning of "If n=c the probability is practically independent of n"? What does independent mean in this case? And how do we obtain e^-1? Thanks! If anyone know a book as "probability and statistics for dummies" please let me know! Thanks ;)

Welcome to MHB, alfred! :)You seem to have left out a couple of parentheses.
The probability is $$\frac {(n-1)^c}{n^c} = \left(\frac {n-1}{n}\right)^c = \left(1 - \frac {1}{n}\right)^c$$.If n=c, the probability becomes $$\left(1 - \frac {1}{n}\right)^n$$.
If n is large enough this will approximate its limit for $n \to \infty$.Since $$\frac 1 e=\lim_{n \to \infty} \left(1-\frac 1 n\right)^n$$ (see for instance wiki), the probability approximates $e^{-1}$.