finding the probability of an event, not equiprobable cases.












1












$begingroup$


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







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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
    16 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
    16 hours ago










  • $begingroup$
    Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
    $endgroup$
    – Henry
    9 hours ago


















1












$begingroup$


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







share|cite|improve this question









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
    16 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
    16 hours ago










  • $begingroup$
    Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
    $endgroup$
    – Henry
    9 hours ago
















1












1








1





$begingroup$


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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 16 hours ago







Ala Bab













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 16 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
    16 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
    16 hours ago










  • $begingroup$
    Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
    $endgroup$
    – Henry
    9 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
    16 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
    16 hours ago










  • $begingroup$
    Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
    $endgroup$
    – Henry
    9 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
16 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
16 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
16 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
16 hours ago












$begingroup$
Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
$endgroup$
– Henry
9 hours ago






$begingroup$
Ala Bab : $(6 times 5 times 4 times 3) + {4 choose 1} times 3 times ( 6 times 5 times 4) = 1800$
$endgroup$
– Henry
9 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

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).






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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.



    $$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      4












      $begingroup$

      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).






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        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).






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          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).






          share|cite|improve this answer









          $endgroup$



          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).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 15 hours ago









          IngixIngix

          4,357149




          4,357149























              3












              $begingroup$

              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.



              $$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                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.



                $$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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.



                  $$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$






                  share|cite|improve this answer









                  $endgroup$



                  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.



                  $$4cdot frac13cdot frac69cdot frac59cdot frac49+frac69cdot frac59cdot frac49cdot frac39=frac{1800}{9^4}approx 0.27.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 15 hours ago









                  farruhotafarruhota

                  20.4k2739




                  20.4k2739






















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