Mixing
A party “A” has majority of 60%, B has 30% and C has 10%
$begingroup$
I just need someone to check my work.
A party "A" has majority of 60%, "B" has 30% and "C" has 10%.
A commission is made up by extracting 6 people randomly. I want to calculate the probability that the three parties are represented evenly.
This is what I have done: $$ frac{60}{100} frac{59}{99} frac{30}{98} frac{29}{97} frac{10}{96} frac{9}{95}$$
My reasoning is that I don't care about how they are extracted, but only about their number, so at first I get 2 from A, then 2 from B and then 2 from C. Is this right?
A closely related questions is about finding the probability that the commission contains at least 3 people from A. Then I just did this:
$$ frac{60}{100} frac{59}{99} frac{58}{98} $$
Is my reasoning right?
probability
asked Jan 24 at 10:06
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
I just need someone to check my work.
A party "A" has majority of 60%, "B" has 30% and "C" has 10%.
A commission is made up by extracting 6 people randomly. I want to calculate the probability that the three parties are represented evenly.
This is what I have done: $$ frac{60}{100} frac{59}{99} frac{30}{98} frac{29}{97} frac{10}{96} frac{9}{95}$$
My reasoning is that I don't care about how they are extracted, but only about their number, so at first I get 2 from A, then 2 from B and then 2 from C. Is this right?
A closely related questions is about finding the probability that the commission contains at least 3 people from A. Then I just did this:
$$ frac{60}{100} frac{59}{99} frac{58}{98} $$
Is my reasoning right?
probability
asked Jan 24 at 10:06
qcc101qcc101
627213
$endgroup$
$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19
add a comment |
$begingroup$
I just need someone to check my work.
A party "A" has majority of 60%, "B" has 30% and "C" has 10%.
A commission is made up by extracting 6 people randomly. I want to calculate the probability that the three parties are represented evenly.
This is what I have done: $$ frac{60}{100} frac{59}{99} frac{30}{98} frac{29}{97} frac{10}{96} frac{9}{95}$$
My reasoning is that I don't care about how they are extracted, but only about their number, so at first I get 2 from A, then 2 from B and then 2 from C. Is this right?
A closely related questions is about finding the probability that the commission contains at least 3 people from A. Then I just did this:
$$ frac{60}{100} frac{59}{99} frac{58}{98} $$
Is my reasoning right?
probability
asked Jan 24 at 10:06
qcc101qcc101
627213
$endgroup$
I just need someone to check my work.
A party "A" has majority of 60%, "B" has 30% and "C" has 10%.
A commission is made up by extracting 6 people randomly. I want to calculate the probability that the three parties are represented evenly.
This is what I have done: $$ frac{60}{100} frac{59}{99} frac{30}{98} frac{29}{97} frac{10}{96} frac{9}{95}$$
My reasoning is that I don't care about how they are extracted, but only about their number, so at first I get 2 from A, then 2 from B and then 2 from C. Is this right?
A closely related questions is about finding the probability that the commission contains at least 3 people from A. Then I just did this:
$$ frac{60}{100} frac{59}{99} frac{58}{98} $$
Is my reasoning right?
probability
probability
asked Jan 24 at 10:06
qcc101qcc101
627213
asked Jan 24 at 10:06
qcc101qcc101
627213
asked Jan 24 at 10:06
qcc101qcc101
627213
asked Jan 24 at 10:06
qcc101qcc101
627213
asked Jan 24 at 10:06
qcc101qcc101
627213
627213
$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19
add a comment |
$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19
$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer equals the probability that - if there are $100$ party members exactly and the committee members are chosen one by one - we end up with the result $AABBCC$.
But there are possibilities (e.g. $ABACCB$) that give evenly presentation.
These possilities are equiprobable and mutually exclusive so to be found is: how much possibilities are there?
The answer is $frac{6!}{2!2!2!}$ and with that factor your answer must be multiplied in order to get the correct answer.
This all under the restriction that there are exactly $100$ party members.
It is not unthinkable that there are e.g. only $10$ or e.g. a whole lot more.
answered Jan 24 at 10:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Your answer equals the probability that - if there are $100$ party members exactly and the committee members are chosen one by one - we end up with the result $AABBCC$.
But there are possibilities (e.g. $ABACCB$) that give evenly presentation.
These possilities are equiprobable and mutually exclusive so to be found is: how much possibilities are there?
The answer is $frac{6!}{2!2!2!}$ and with that factor your answer must be multiplied in order to get the correct answer.
This all under the restriction that there are exactly $100$ party members.
It is not unthinkable that there are e.g. only $10$ or e.g. a whole lot more.
answered Jan 24 at 10:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
add a comment |
$begingroup$
Your answer equals the probability that - if there are $100$ party members exactly and the committee members are chosen one by one - we end up with the result $AABBCC$.
But there are possibilities (e.g. $ABACCB$) that give evenly presentation.
These possilities are equiprobable and mutually exclusive so to be found is: how much possibilities are there?
The answer is $frac{6!}{2!2!2!}$ and with that factor your answer must be multiplied in order to get the correct answer.
This all under the restriction that there are exactly $100$ party members.
It is not unthinkable that there are e.g. only $10$ or e.g. a whole lot more.
answered Jan 24 at 10:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
add a comment |
$begingroup$
Your answer equals the probability that - if there are $100$ party members exactly and the committee members are chosen one by one - we end up with the result $AABBCC$.
But there are possibilities (e.g. $ABACCB$) that give evenly presentation.
These possilities are equiprobable and mutually exclusive so to be found is: how much possibilities are there?
The answer is $frac{6!}{2!2!2!}$ and with that factor your answer must be multiplied in order to get the correct answer.
This all under the restriction that there are exactly $100$ party members.
It is not unthinkable that there are e.g. only $10$ or e.g. a whole lot more.
answered Jan 24 at 10:19
drhabdrhab
103k545136
$endgroup$
Your answer equals the probability that - if there are $100$ party members exactly and the committee members are chosen one by one - we end up with the result $AABBCC$.
But there are possibilities (e.g. $ABACCB$) that give evenly presentation.
These possilities are equiprobable and mutually exclusive so to be found is: how much possibilities are there?
The answer is $frac{6!}{2!2!2!}$ and with that factor your answer must be multiplied in order to get the correct answer.
This all under the restriction that there are exactly $100$ party members.
It is not unthinkable that there are e.g. only $10$ or e.g. a whole lot more.
answered Jan 24 at 10:19
drhabdrhab
103k545136
answered Jan 24 at 10:19
drhabdrhab
103k545136
answered Jan 24 at 10:19
drhabdrhab
103k545136
answered Jan 24 at 10:19
drhabdrhab
103k545136
103k545136
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
add a comment |
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
I understood the part about the coefficients. However what if I do not know the total number of people?
$endgroup$
– qcc101
Jan 24 at 10:21
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
$begingroup$
If that number is small then without further knowledge you cannot calculate the probability. If that number is big then you can go for almost the same route in order to get a good approximation. The only difference is that your original answer changes into $0.6^20.3^20.1^2$ and the approximation will be $frac{6!}{2!2!2!}0.6^20.3^20.1^2$. You do not take into account then that - if one member of a party has been chosen already - the number of candidates in that party has decreased with $1$. By big numbers that does not harm very much.
$endgroup$
– drhab
Jan 24 at 10:23
add a comment |
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$begingroup$
You havent regarded the order. You have calculated the probability for the order AABBCC, but ABACBC ist possible as well. You have to regard the order.
$endgroup$
– callculus
Jan 24 at 10:12
$begingroup$
@callculus Oh, you are right. Then I just multiply by 6!, is that right?
$endgroup$
– qcc101
Jan 24 at 10:13
$begingroup$
The number of order is not just 6! since you do not distinguish between $A_1$ and $A_2$. The number of ways to arrange A,A,B,B,C,C is $frac{6!}{2!cdot 2!cdot 2!}$
$endgroup$
– callculus
Jan 24 at 10:16
$begingroup$
Got it! Then the second answer changes as well, this time I need to multiply by $frac{6!}{3! 3!}$
$endgroup$
– qcc101
Jan 24 at 10:19