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?
probability
asked Nov 22 at 16:44
Man
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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.
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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
asked Nov 22 at 16:44
Man
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.
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
add a comment |
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
asked Nov 22 at 16:44
Man
162
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
asked Nov 22 at 16:44
Man
162
asked Nov 22 at 16:44
Man
162
asked Nov 22 at 16:44
Man
162
asked Nov 22 at 16:44
Man
162
asked Nov 22 at 16:44
Man
162
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
add a comment |
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
add a comment |
1 Answer
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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
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$
answered Nov 22 at 16:51
Akash Roy
1
Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58
add a comment |
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
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$
answered Nov 22 at 16:51
Akash Roy
1
Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58
add a comment |
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
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$
answered Nov 22 at 16:51
Akash Roy
1
Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58
add a comment |
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
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$
answered Nov 22 at 16:51
Akash Roy
1
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
$$frac{{9choose 3}cdot{6choose 4}}{15choose 7}$$
answered Nov 22 at 16:51
Akash Roy
1
edited Nov 22 at 16:57
answered Nov 22 at 16:51
Akash Roy
1
answered Nov 22 at 16:51
Akash Roy
1
answered Nov 22 at 16:51
Akash Roy
1
1
Thanks @Man pleasure to help you
– Akash Roy
Nov 22 at 17:58
add a comment |
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
add a comment |
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