finding the probability of an event, not equiprobable cases.Another Birthday Problem...
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finding the probability of an event, not equiprobable cases.
Another Birthday Problem (Probability/Combinatorics)The man with two boysAt least how many persons are required in a group so that the probability of two persons were born on the same day of the week is 0.5?Finding the probability for specific birthday problem.finding the probability that Alice has at least one class each dayProbability that both children were born on Mondays given that the older child was born on a MondayOther days of the week - Birthday Paradoxwhat is the probability that in a random group of seven people two were born on Monday and two on Sunday?Can't wrap around, what's wrong with the probability calculation. Please helpProbability and statistics 11
$begingroup$
A person can be born on a Monday with probability 1/3 and on any other
day with equal probability. What is the probability that 4 randomly
chosen persons were born on different days of the week.
6!/4! * (1/9)^4 + 6!/3! * (1/9)^3 * 1/3
denominator = 7^4
*took two cases where one 4 days are possible but not monday
other case is when it might be on a monday and 3 other days
the answer seemed not correct
probability combinatorics permutations
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A person can be born on a Monday with probability 1/3 and on any other
day with equal probability. What is the probability that 4 randomly
chosen persons were born on different days of the week.
6!/4! * (1/9)^4 + 6!/3! * (1/9)^3 * 1/3
denominator = 7^4
*took two cases where one 4 days are possible but not monday
other case is when it might be on a monday and 3 other days
the answer seemed not correct
probability combinatorics permutations
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago
add a comment |
$begingroup$
A person can be born on a Monday with probability 1/3 and on any other
day with equal probability. What is the probability that 4 randomly
chosen persons were born on different days of the week.
6!/4! * (1/9)^4 + 6!/3! * (1/9)^3 * 1/3
denominator = 7^4
*took two cases where one 4 days are possible but not monday
other case is when it might be on a monday and 3 other days
the answer seemed not correct
probability combinatorics permutations
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A person can be born on a Monday with probability 1/3 and on any other
day with equal probability. What is the probability that 4 randomly
chosen persons were born on different days of the week.
6!/4! * (1/9)^4 + 6!/3! * (1/9)^3 * 1/3
denominator = 7^4
*took two cases where one 4 days are possible but not monday
other case is when it might be on a monday and 3 other days
the answer seemed not correct
probability combinatorics permutations
probability combinatorics permutations
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Ala Bab
asked 6 hours ago
Ala BabAla Bab
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Ala BabAla Bab
214
asked 6 hours ago
Ala BabAla Bab
214
214
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ala Bab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago
add a comment |
$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago
$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have
$$6times5times4times3 = {6! over 2!}$$
possibilities to choose the (ordered) Weekdays ($neq$ Monday).
In the 2nd case (one Monday in the set), you have
$$6times5times4 = {6! over 3!}$$
ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.
That means your overall probability is
$${6! over 2!}left({1 over 9}right)^4 + 4times{6! over 3!}left({1 over 9}right)^3{1 over 3} = {360 + 4times120times3 over 9^4}={1800 over 9^4}={200 over 9^3} approx 0.274ldots$$
I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $left({1 over 9}right)^4$ and similar probabilities depend on an order.
To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $left({1 over 9}right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $left({1 over 9}right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $left({1 over 9}right)^4$.
If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $left({1 over 9}right)^4$, this is too small by a factor of 24.
EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).
answered 6 hours ago
IngixIngix
4,357149
$endgroup$
add a comment |
$begingroup$
Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.
$$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$
answered 6 hours ago
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have
$$6times5times4times3 = {6! over 2!}$$
possibilities to choose the (ordered) Weekdays ($neq$ Monday).
In the 2nd case (one Monday in the set), you have
$$6times5times4 = {6! over 3!}$$
ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.
That means your overall probability is
$${6! over 2!}left({1 over 9}right)^4 + 4times{6! over 3!}left({1 over 9}right)^3{1 over 3} = {360 + 4times120times3 over 9^4}={1800 over 9^4}={200 over 9^3} approx 0.274ldots$$
I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $left({1 over 9}right)^4$ and similar probabilities depend on an order.
To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $left({1 over 9}right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $left({1 over 9}right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $left({1 over 9}right)^4$.
If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $left({1 over 9}right)^4$, this is too small by a factor of 24.
EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).
answered 6 hours ago
IngixIngix
4,357149
$endgroup$
add a comment |
$begingroup$
You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have
$$6times5times4times3 = {6! over 2!}$$
possibilities to choose the (ordered) Weekdays ($neq$ Monday).
In the 2nd case (one Monday in the set), you have
$$6times5times4 = {6! over 3!}$$
ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.
That means your overall probability is
$${6! over 2!}left({1 over 9}right)^4 + 4times{6! over 3!}left({1 over 9}right)^3{1 over 3} = {360 + 4times120times3 over 9^4}={1800 over 9^4}={200 over 9^3} approx 0.274ldots$$
I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $left({1 over 9}right)^4$ and similar probabilities depend on an order.
To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $left({1 over 9}right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $left({1 over 9}right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $left({1 over 9}right)^4$.
If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $left({1 over 9}right)^4$, this is too small by a factor of 24.
EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).
answered 6 hours ago
IngixIngix
4,357149
$endgroup$
add a comment |
$begingroup$
You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have
$$6times5times4times3 = {6! over 2!}$$
possibilities to choose the (ordered) Weekdays ($neq$ Monday).
In the 2nd case (one Monday in the set), you have
$$6times5times4 = {6! over 3!}$$
ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.
That means your overall probability is
$${6! over 2!}left({1 over 9}right)^4 + 4times{6! over 3!}left({1 over 9}right)^3{1 over 3} = {360 + 4times120times3 over 9^4}={1800 over 9^4}={200 over 9^3} approx 0.274ldots$$
I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $left({1 over 9}right)^4$ and similar probabilities depend on an order.
To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $left({1 over 9}right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $left({1 over 9}right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $left({1 over 9}right)^4$.
If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $left({1 over 9}right)^4$, this is too small by a factor of 24.
EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).
answered 6 hours ago
IngixIngix
4,357149
$endgroup$
You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have
$$6times5times4times3 = {6! over 2!}$$
possibilities to choose the (ordered) Weekdays ($neq$ Monday).
In the 2nd case (one Monday in the set), you have
$$6times5times4 = {6! over 3!}$$
ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.
That means your overall probability is
$${6! over 2!}left({1 over 9}right)^4 + 4times{6! over 3!}left({1 over 9}right)^3{1 over 3} = {360 + 4times120times3 over 9^4}={1800 over 9^4}={200 over 9^3} approx 0.274ldots$$
I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $left({1 over 9}right)^4$ and similar probabilities depend on an order.
To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $left({1 over 9}right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $left({1 over 9}right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $left({1 over 9}right)^4$.
If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $left({1 over 9}right)^4$, this is too small by a factor of 24.
EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).
answered 6 hours ago
IngixIngix
4,357149
answered 6 hours ago
IngixIngix
4,357149
answered 6 hours ago
IngixIngix
4,357149
answered 6 hours ago
IngixIngix
4,357149
4,357149
add a comment |
add a comment |
$begingroup$
Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.
$$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$
answered 6 hours ago
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.
$$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$
answered 6 hours ago
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.
$$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$
answered 6 hours ago
farruhotafarruhota
20.4k2739
$endgroup$
Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.
$$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$
answered 6 hours ago
farruhotafarruhota
20.4k2739
answered 6 hours ago
farruhotafarruhota
20.4k2739
answered 6 hours ago
farruhotafarruhota
20.4k2739
answered 6 hours ago
farruhotafarruhota
20.4k2739
20.4k2739
add a comment |
add a comment |
Ala Bab is a new contributor. Be nice, and check out our Code of Conduct.
Ala Bab is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $frac19$s in the numerator? My guess is that the answer is $frac{1800}{9^4}approx 0.2743$
$endgroup$
– Henry
6 hours ago
$begingroup$
@Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday?
$endgroup$
– Ala Bab
6 hours ago