probability of a board containing a certain number of accountants
A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?
cant figure this one out
probability statistics
asked Apr 14 '13 at 0:08
notamathwiz
50341330
add a comment |
A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?
cant figure this one out
probability statistics
asked Apr 14 '13 at 0:08
notamathwiz
50341330
2
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11
add a comment |
A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?
cant figure this one out
probability statistics
asked Apr 14 '13 at 0:08
notamathwiz
50341330
A corporate board contains twelve members. The board decides to create a five-person Committee
to Hide Corporation Debt. Suppose four members of the board are accountants. What is the probability that the Committee will contain two accountants and three nonaccountants?
cant figure this one out
probability statistics
probability statistics
asked Apr 14 '13 at 0:08
notamathwiz
50341330
asked Apr 14 '13 at 0:08
notamathwiz
50341330
asked Apr 14 '13 at 0:08
notamathwiz
50341330
asked Apr 14 '13 at 0:08
notamathwiz
50341330
asked Apr 14 '13 at 0:08
notamathwiz
50341330
50341330
2
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11
add a comment |
2
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11
2
2
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11
add a comment |
2 Answers
2
active
oldest
votes
There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.
But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
|
show 1 more comment
Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$
$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$
edited yesterday
Davide Giraudo
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.
But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
|
show 1 more comment
There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.
But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
|
show 1 more comment
There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.
But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
There are $binom{12}{5}$ ways to choose the $5$ people from the $12$. We are expected to assume these choices are all equally likely. That is a very unreasonable assumption. It is simply not true that committees are chosen by tossing coins or dice.
But let's hold our noses and go on. Call a choice good if it consists of $2$ accountants and $3$ others. There are $binom{4}{2}$ ways to choose the $2$ accountants. For every such choice, there are $binom{8}{3}$ ways to choose the $3$ others. So how many good choices are there?
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
answered Apr 14 '13 at 0:27
André Nicolas
451k36422806
451k36422806
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
|
show 1 more comment
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
so the good choices would be $$ frac{binom{4}{2} binom{8}{3}}{binom{12}{5}} $$ would this be correct?
– notamathwiz
Apr 14 '13 at 0:34
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
You have written down the probability correctly. The number of good choices is $binom{4}{2}binom{8}{3}$.
– André Nicolas
Apr 14 '13 at 0:37
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
this would equal to 62 / 792 which is 0.08 which is the probability that the Committee will contain two accountants and three nonaccountants? can someone concur
– notamathwiz
Apr 14 '13 at 0:40
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
The denominator is right. The numerator isn't.
– André Nicolas
Apr 14 '13 at 1:09
1
1
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
Yes, that's right. In your first comment, you correctly wrote that one should multiply $binom{4}{2}$ and $binom{8}{3}$. But when calculating, you pressed the $+$ button. It is safer to do it in one's head.
– André Nicolas
Apr 14 '13 at 1:37
|
show 1 more comment
Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$
$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$
edited yesterday
Davide Giraudo
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$
$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$
edited yesterday
Davide Giraudo
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$
$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$
edited yesterday
Davide Giraudo
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Using the Hypergeometric distribution formular,
$$p(x) = {{k choose x} {N-k choose n-x} over {N choose n} }$$
$N=12$ , $n=5$ ,
$k=4$ ,
$x=2$ ,
$n-x=3$ (non-accountants) , $p(x) = 14/33$
edited yesterday
Davide Giraudo
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Davide Giraudo
125k16150260
edited yesterday
Davide Giraudo
125k16150260
edited yesterday
Davide Giraudo
125k16150260
125k16150260
answered yesterday
Janet Anagli
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Janet Anagli
1
answered yesterday
Janet Anagli
1
1
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Janet Anagli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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2
binomial distribution? why are you asking 2 questions in 1 minute on the same topic rather than wait for a hint on one of them and see if you can do the other on your own. this is homework right?
– Lost1
Apr 14 '13 at 0:11