Elementary probability theory [duplicate]











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  • Probability of sampling with and without replacement

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I would like to ask a simple question but I'm unable to understand the logic behind it.




From an urn containing $M$ balls $n$ balls are drawn with replacement. What is the probability that at least one ball is drawn more than once?




Thank you in advance.






probability-theory







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marked as duplicate by caverac, Davide Giraudo probability-theory
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 Nov 21 at 16:33


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  • I checked the possible duplicate question but seems like the other one is without replacement.
    – M K
    Nov 21 at 15:17










  • There're two sections: with replacement and without replacement
    – caverac
    Nov 21 at 15:18

















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  • Probability of sampling with and without replacement

    2 answers




I would like to ask a simple question but I'm unable to understand the logic behind it.




From an urn containing $M$ balls $n$ balls are drawn with replacement. What is the probability that at least one ball is drawn more than once?




Thank you in advance.






probability-theory







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marked as duplicate by caverac, Davide Giraudo probability-theory
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 Nov 21 at 16:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • I checked the possible duplicate question but seems like the other one is without replacement.
    – M K
    Nov 21 at 15:17










  • There're two sections: with replacement and without replacement
    – caverac
    Nov 21 at 15:18















up vote
0
down vote

favorite
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up vote
0
down vote

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1






This question already has an answer here:




  • Probability of sampling with and without replacement

    2 answers




I would like to ask a simple question but I'm unable to understand the logic behind it.




From an urn containing $M$ balls $n$ balls are drawn with replacement. What is the probability that at least one ball is drawn more than once?




Thank you in advance.






probability-theory







share|cite|improve this question
















This question already has an answer here:




  • Probability of sampling with and without replacement

    2 answers




I would like to ask a simple question but I'm unable to understand the logic behind it.




From an urn containing $M$ balls $n$ balls are drawn with replacement. What is the probability that at least one ball is drawn more than once?




Thank you in advance.





This question already has an answer here:




  • Probability of sampling with and without replacement

    2 answers







probability-theory




probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 15:14









caverac

12.5k21027




12.5k21027










asked Nov 21 at 15:12









M K

31




31




marked as duplicate by caverac, Davide Giraudo probability-theory
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 Nov 21 at 16:33


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • I checked the possible duplicate question but seems like the other one is without replacement.
    – M K
    Nov 21 at 15:17










  • There're two sections: with replacement and without replacement
    – caverac
    Nov 21 at 15:18




















  • I checked the possible duplicate question but seems like the other one is without replacement.
    – M K
    Nov 21 at 15:17










  • There're two sections: with replacement and without replacement
    – caverac
    Nov 21 at 15:18


















I checked the possible duplicate question but seems like the other one is without replacement.
– M K
Nov 21 at 15:17




I checked the possible duplicate question but seems like the other one is without replacement.
– M K
Nov 21 at 15:17












There're two sections: with replacement and without replacement
– caverac
Nov 21 at 15:18






There're two sections: with replacement and without replacement
– caverac
Nov 21 at 15:18












2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










In questions such as this, it is useful to think of the complement of these events. What is the complement of at least one ball being drawn more than once; each ball drawn is distinct. So even though you are replacing the balls, the number of 'good' choices starts decreasing. So the first time, you have $M$ choices, the next time $M-1$, and so on. Meanwhile, the total number of possible choices is $M^n$. Does that seem fine?






share|cite|improve this answer





















  • Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
    – M K
    Nov 21 at 15:26










  • How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
    – Boshu
    Nov 21 at 15:35










  • Thank you very much that was helpful!
    – M K
    Nov 21 at 15:48


















up vote
0
down vote













Hint:



The complement is ‘each ball is distinct’. Its cardinal is equal to the number of injections from a set with $n$ elements to a set with $M$ elements.






share|cite|improve this answer























  • Thank you very much. Now I got it!
    – M K
    Nov 21 at 15:49


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










In questions such as this, it is useful to think of the complement of these events. What is the complement of at least one ball being drawn more than once; each ball drawn is distinct. So even though you are replacing the balls, the number of 'good' choices starts decreasing. So the first time, you have $M$ choices, the next time $M-1$, and so on. Meanwhile, the total number of possible choices is $M^n$. Does that seem fine?






share|cite|improve this answer





















  • Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
    – M K
    Nov 21 at 15:26










  • How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
    – Boshu
    Nov 21 at 15:35










  • Thank you very much that was helpful!
    – M K
    Nov 21 at 15:48















up vote
0
down vote



accepted










In questions such as this, it is useful to think of the complement of these events. What is the complement of at least one ball being drawn more than once; each ball drawn is distinct. So even though you are replacing the balls, the number of 'good' choices starts decreasing. So the first time, you have $M$ choices, the next time $M-1$, and so on. Meanwhile, the total number of possible choices is $M^n$. Does that seem fine?






share|cite|improve this answer





















  • Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
    – M K
    Nov 21 at 15:26










  • How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
    – Boshu
    Nov 21 at 15:35










  • Thank you very much that was helpful!
    – M K
    Nov 21 at 15:48













up vote
0
down vote



accepted







up vote
0
down vote



accepted






In questions such as this, it is useful to think of the complement of these events. What is the complement of at least one ball being drawn more than once; each ball drawn is distinct. So even though you are replacing the balls, the number of 'good' choices starts decreasing. So the first time, you have $M$ choices, the next time $M-1$, and so on. Meanwhile, the total number of possible choices is $M^n$. Does that seem fine?






share|cite|improve this answer












In questions such as this, it is useful to think of the complement of these events. What is the complement of at least one ball being drawn more than once; each ball drawn is distinct. So even though you are replacing the balls, the number of 'good' choices starts decreasing. So the first time, you have $M$ choices, the next time $M-1$, and so on. Meanwhile, the total number of possible choices is $M^n$. Does that seem fine?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 15:21









Boshu

698315




698315












  • Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
    – M K
    Nov 21 at 15:26










  • How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
    – Boshu
    Nov 21 at 15:35










  • Thank you very much that was helpful!
    – M K
    Nov 21 at 15:48


















  • Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
    – M K
    Nov 21 at 15:26










  • How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
    – Boshu
    Nov 21 at 15:35










  • Thank you very much that was helpful!
    – M K
    Nov 21 at 15:48
















Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
– M K
Nov 21 at 15:26




Thank you very much for the reply. I understood the logic with your help but the answer is supposed to be P(A) = 1- M*(M-1)....(M-n+1)/M^n as you said. However, I didn't quite get where this (M-n+1) comes from. I'm unable to visualize the answer.
– M K
Nov 21 at 15:26












How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
– Boshu
Nov 21 at 15:35




How many times do you pick the balls? $n$ times. So how many valid options do you have on the last turn?
– Boshu
Nov 21 at 15:35












Thank you very much that was helpful!
– M K
Nov 21 at 15:48




Thank you very much that was helpful!
– M K
Nov 21 at 15:48










up vote
0
down vote













Hint:



The complement is ‘each ball is distinct’. Its cardinal is equal to the number of injections from a set with $n$ elements to a set with $M$ elements.






share|cite|improve this answer























  • Thank you very much. Now I got it!
    – M K
    Nov 21 at 15:49















up vote
0
down vote













Hint:



The complement is ‘each ball is distinct’. Its cardinal is equal to the number of injections from a set with $n$ elements to a set with $M$ elements.






share|cite|improve this answer























  • Thank you very much. Now I got it!
    – M K
    Nov 21 at 15:49













up vote
0
down vote










up vote
0
down vote









Hint:



The complement is ‘each ball is distinct’. Its cardinal is equal to the number of injections from a set with $n$ elements to a set with $M$ elements.






share|cite|improve this answer














Hint:



The complement is ‘each ball is distinct’. Its cardinal is equal to the number of injections from a set with $n$ elements to a set with $M$ elements.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 16:03

























answered Nov 21 at 15:36









Bernard

117k637109




117k637109












  • Thank you very much. Now I got it!
    – M K
    Nov 21 at 15:49


















  • Thank you very much. Now I got it!
    – M K
    Nov 21 at 15:49
















Thank you very much. Now I got it!
– M K
Nov 21 at 15:49




Thank you very much. Now I got it!
– M K
Nov 21 at 15:49



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