Probability that the number of red balls removed from the bag is 4 [closed]











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A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6 balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from the bag. What is the probability that the number of red balls removed from the bag is exactly 4?






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closed as off-topic by JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo Nov 23 at 1:16


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo

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  • en.wikipedia.org/wiki/Hypergeometric_distribution
    – JMoravitz
    Nov 22 at 16:45










  • @Moravitz please check if my answer is correct.
    – Akash Roy
    Nov 22 at 16:53















up vote
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down vote

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A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6 balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from the bag. What is the probability that the number of red balls removed from the bag is exactly 4?






probability







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closed as off-topic by JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo Nov 23 at 1:16


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • en.wikipedia.org/wiki/Hypergeometric_distribution
    – JMoravitz
    Nov 22 at 16:45










  • @Moravitz please check if my answer is correct.
    – Akash Roy
    Nov 22 at 16:53













up vote
0
down vote

favorite









up vote
0
down vote

favorite











A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6 balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from the bag. What is the probability that the number of red balls removed from the bag is exactly 4?






probability







share|cite|improve this question













A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6 balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from the bag. What is the probability that the number of red balls removed from the bag is exactly 4?






probability




probability






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share|cite|improve this question











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asked Nov 22 at 16:44









Man

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162




closed as off-topic by JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo Nov 23 at 1:16


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo Nov 23 at 1:16


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, Rebellos, Davide Giraudo, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • en.wikipedia.org/wiki/Hypergeometric_distribution
    – JMoravitz
    Nov 22 at 16:45










  • @Moravitz please check if my answer is correct.
    – Akash Roy
    Nov 22 at 16:53


















  • en.wikipedia.org/wiki/Hypergeometric_distribution
    – JMoravitz
    Nov 22 at 16:45










  • @Moravitz please check if my answer is correct.
    – Akash Roy
    Nov 22 at 16:53
















en.wikipedia.org/wiki/Hypergeometric_distribution
– JMoravitz
Nov 22 at 16:45




en.wikipedia.org/wiki/Hypergeometric_distribution
– JMoravitz
Nov 22 at 16:45












@Moravitz please check if my answer is correct.
– Akash Roy
Nov 22 at 16:53




@Moravitz please check if my answer is correct.
– Akash Roy
Nov 22 at 16:53










1 Answer
1






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oldest

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0
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accepted










There are $15choose 7$ ways to make the required selections. Of these, there are ${9choose 3}cdot {6choose 4}$ ways to choose exactly $4$ red balls.



Therefore your solution should be
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$






share|cite|improve this answer























  • Thanks @Man pleasure to help you
    – Akash Roy
    Nov 22 at 17:58


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










There are $15choose 7$ ways to make the required selections. Of these, there are ${9choose 3}cdot {6choose 4}$ ways to choose exactly $4$ red balls.



Therefore your solution should be
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$






share|cite|improve this answer























  • Thanks @Man pleasure to help you
    – Akash Roy
    Nov 22 at 17:58















up vote
0
down vote



accepted










There are $15choose 7$ ways to make the required selections. Of these, there are ${9choose 3}cdot {6choose 4}$ ways to choose exactly $4$ red balls.



Therefore your solution should be
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$






share|cite|improve this answer























  • Thanks @Man pleasure to help you
    – Akash Roy
    Nov 22 at 17:58













up vote
0
down vote



accepted







up vote
0
down vote



accepted






There are $15choose 7$ ways to make the required selections. Of these, there are ${9choose 3}cdot {6choose 4}$ ways to choose exactly $4$ red balls.



Therefore your solution should be
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$






share|cite|improve this answer














There are $15choose 7$ ways to make the required selections. Of these, there are ${9choose 3}cdot {6choose 4}$ ways to choose exactly $4$ red balls.



Therefore your solution should be
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 at 16:57

























answered Nov 22 at 16:51









Akash Roy

1




1












  • Thanks @Man pleasure to help you
    – Akash Roy
    Nov 22 at 17:58


















  • Thanks @Man pleasure to help you
    – Akash Roy
    Nov 22 at 17:58
















Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58




Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58



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