Probability of Not Getting Candy

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In summary, the probability of my nephew not receiving any candy in a party where c candies are given randomly to n kids is (n-1/n)^c, and this value approximates to e^-1 when n is large enough. "Independent" in this case means that the probability does not depend on the value of n. The value of e^-1 is obtained through the limit \frac 1 e=\lim_{n \to \infty} \left(1-\frac 1 n\right)^n.
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alfred2
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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 ;)
 
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  • #2
alfred said:
Hi I've found this exercise but I do not understand the solution:
_________________________________________________________________
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 \(\displaystyle \frac {(n-1)^c}{n^c} = \left(\frac {n-1}{n}\right)^c = \left(1 - \frac {1}{n}\right)^c\).If n=c, the probability becomes \(\displaystyle \left(1 - \frac {1}{n}\right)^n\).
If n is large enough this will approximate its limit for $n \to \infty$.Since \(\displaystyle \frac 1 e=\lim_{n \to \infty} \left(1-\frac 1 n\right)^n\) (see for instance wiki), the probability approximates $e^{-1}$.
 

FAQ: Probability of Not Getting Candy

What is the probability of not getting candy?

The probability of not getting candy depends on the specific situation and the factors involved. It can range from 0% (if all factors indicate that candy is certain) to 100% (if all factors indicate that candy is impossible).

How is the probability of not getting candy calculated?

The probability of not getting candy is calculated by dividing the number of outcomes where candy is not received by the total number of possible outcomes. This can be expressed as a fraction, decimal, or percentage.

What factors affect the probability of not getting candy?

Some factors that may affect the probability of not getting candy include the number of people in a group, the amount of candy available, and the distribution method (e.g. random selection or predetermined). Other factors such as preferences and personal behavior may also play a role.

Can the probability of not getting candy change?

Yes, the probability of not getting candy can change depending on the situation. For example, if more candy is added to the available pool, the probability of not getting candy decreases. Similarly, if more people join the group, the probability of not getting candy may increase.

Why is it important to understand the probability of not getting candy?

Understanding the probability of not getting candy can help manage expectations and avoid disappointment. It can also be useful in decision making, such as determining how much candy to purchase or how to distribute it fairly among a group. Additionally, understanding probability in general can help with critical thinking and problem solving skills.

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