Calculating Probability using Bayes Theorem

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3












$begingroup$


I'm trying to calculate using the bayes theorem the exercise below.



But Im confused.



In one company, 40% of the employees are women. Suppose 60% of men workers are married, and 40% of women workers are married. How likely is a married worker to be a man?



What is the correct answer? I found about 0.69.



36/52. that's right?






probability bayes-theorem







share|cite|improve this question









$endgroup$











  • $begingroup$
    Yes, that answer looks good.
    $endgroup$
    – lulu
    Jan 13 at 22:12










  • $begingroup$
    @lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
    $endgroup$
    – Laura
    Jan 13 at 22:17
















3












$begingroup$


I'm trying to calculate using the bayes theorem the exercise below.



But Im confused.



In one company, 40% of the employees are women. Suppose 60% of men workers are married, and 40% of women workers are married. How likely is a married worker to be a man?



What is the correct answer? I found about 0.69.



36/52. that's right?






probability bayes-theorem







share|cite|improve this question









$endgroup$











  • $begingroup$
    Yes, that answer looks good.
    $endgroup$
    – lulu
    Jan 13 at 22:12










  • $begingroup$
    @lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
    $endgroup$
    – Laura
    Jan 13 at 22:17














3












3








3





$begingroup$


I'm trying to calculate using the bayes theorem the exercise below.



But Im confused.



In one company, 40% of the employees are women. Suppose 60% of men workers are married, and 40% of women workers are married. How likely is a married worker to be a man?



What is the correct answer? I found about 0.69.



36/52. that's right?






probability bayes-theorem







share|cite|improve this question









$endgroup$




I'm trying to calculate using the bayes theorem the exercise below.



But Im confused.



In one company, 40% of the employees are women. Suppose 60% of men workers are married, and 40% of women workers are married. How likely is a married worker to be a man?



What is the correct answer? I found about 0.69.



36/52. that's right?






probability bayes-theorem




probability bayes-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 at 22:08









LauraLaura

2288




2288











  • $begingroup$
    Yes, that answer looks good.
    $endgroup$
    – lulu
    Jan 13 at 22:12










  • $begingroup$
    @lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
    $endgroup$
    – Laura
    Jan 13 at 22:17

















  • $begingroup$
    Yes, that answer looks good.
    $endgroup$
    – lulu
    Jan 13 at 22:12










  • $begingroup$
    @lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
    $endgroup$
    – Laura
    Jan 13 at 22:17
















$begingroup$
Yes, that answer looks good.
$endgroup$
– lulu
Jan 13 at 22:12




$begingroup$
Yes, that answer looks good.
$endgroup$
– lulu
Jan 13 at 22:12












$begingroup$
@lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
$endgroup$
– Laura
Jan 13 at 22:17





$begingroup$
@lulu how can I use the Bayes Theorem? And I believe this is not the right answer.
$endgroup$
– Laura
Jan 13 at 22:17











2 Answers
2






active

oldest

votes


















4












$begingroup$

Indeed:



$P(man|married) = fracP(married/man) cdot P(man)P(married/man) cdot P(man) + P(married/woman) cdot P(woman) = frac0.6 cdot 0.60.6 cdot 0.6 + 0.4 cdot 0.4=frac913$



So you are right!






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Thanks @callculus for pointing that out. Corrected!
    $endgroup$
    – pendermath
    Jan 13 at 22:24






  • 1




    $begingroup$
    Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
    $endgroup$
    – pendermath
    Jan 13 at 22:25






  • 1




    $begingroup$
    @pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
    $endgroup$
    – callculus
    Jan 13 at 22:26







  • 2




    $begingroup$
    The solution of question 1941655 is also $9/13$. Am I missing something?
    $endgroup$
    – pendermath
    Jan 13 at 22:35






  • 2




    $begingroup$
    @Laura But a nice answer.
    $endgroup$
    – callculus
    Jan 13 at 22:47


















3












$begingroup$

You can also do this with an example. Suppose there are $100$ employees. Then there are $60$ men, $0.6cdot60=36$ of whom are married, and there are $40$ women, $0.4cdot40=16$ of whom are married. There are then $36+16=52$ married workers, $36$ of whom are men, so the probability in question is $36over52=9over13$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Sure, it's just that @Laura wanted it solved using Bayes' Theorem
    $endgroup$
    – pendermath
    Jan 13 at 22:30










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Indeed:



$P(man|married) = fracP(married/man) cdot P(man)P(married/man) cdot P(man) + P(married/woman) cdot P(woman) = frac0.6 cdot 0.60.6 cdot 0.6 + 0.4 cdot 0.4=frac913$



So you are right!






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Thanks @callculus for pointing that out. Corrected!
    $endgroup$
    – pendermath
    Jan 13 at 22:24






  • 1




    $begingroup$
    Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
    $endgroup$
    – pendermath
    Jan 13 at 22:25






  • 1




    $begingroup$
    @pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
    $endgroup$
    – callculus
    Jan 13 at 22:26







  • 2




    $begingroup$
    The solution of question 1941655 is also $9/13$. Am I missing something?
    $endgroup$
    – pendermath
    Jan 13 at 22:35






  • 2




    $begingroup$
    @Laura But a nice answer.
    $endgroup$
    – callculus
    Jan 13 at 22:47















4












$begingroup$

Indeed:



$P(man|married) = fracP(married/man) cdot P(man)P(married/man) cdot P(man) + P(married/woman) cdot P(woman) = frac0.6 cdot 0.60.6 cdot 0.6 + 0.4 cdot 0.4=frac913$



So you are right!






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Thanks @callculus for pointing that out. Corrected!
    $endgroup$
    – pendermath
    Jan 13 at 22:24






  • 1




    $begingroup$
    Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
    $endgroup$
    – pendermath
    Jan 13 at 22:25






  • 1




    $begingroup$
    @pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
    $endgroup$
    – callculus
    Jan 13 at 22:26







  • 2




    $begingroup$
    The solution of question 1941655 is also $9/13$. Am I missing something?
    $endgroup$
    – pendermath
    Jan 13 at 22:35






  • 2




    $begingroup$
    @Laura But a nice answer.
    $endgroup$
    – callculus
    Jan 13 at 22:47













4












4








4





$begingroup$

Indeed:



$P(man|married) = fracP(married/man) cdot P(man)P(married/man) cdot P(man) + P(married/woman) cdot P(woman) = frac0.6 cdot 0.60.6 cdot 0.6 + 0.4 cdot 0.4=frac913$



So you are right!






share|cite|improve this answer











$endgroup$



Indeed:



$P(man|married) = fracP(married/man) cdot P(man)P(married/man) cdot P(man) + P(married/woman) cdot P(woman) = frac0.6 cdot 0.60.6 cdot 0.6 + 0.4 cdot 0.4=frac913$



So you are right!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 22:22

























answered Jan 13 at 22:19









pendermathpendermath

54410




54410







  • 1




    $begingroup$
    Thanks @callculus for pointing that out. Corrected!
    $endgroup$
    – pendermath
    Jan 13 at 22:24






  • 1




    $begingroup$
    Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
    $endgroup$
    – pendermath
    Jan 13 at 22:25






  • 1




    $begingroup$
    @pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
    $endgroup$
    – callculus
    Jan 13 at 22:26







  • 2




    $begingroup$
    The solution of question 1941655 is also $9/13$. Am I missing something?
    $endgroup$
    – pendermath
    Jan 13 at 22:35






  • 2




    $begingroup$
    @Laura But a nice answer.
    $endgroup$
    – callculus
    Jan 13 at 22:47












  • 1




    $begingroup$
    Thanks @callculus for pointing that out. Corrected!
    $endgroup$
    – pendermath
    Jan 13 at 22:24






  • 1




    $begingroup$
    Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
    $endgroup$
    – pendermath
    Jan 13 at 22:25






  • 1




    $begingroup$
    @pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
    $endgroup$
    – callculus
    Jan 13 at 22:26







  • 2




    $begingroup$
    The solution of question 1941655 is also $9/13$. Am I missing something?
    $endgroup$
    – pendermath
    Jan 13 at 22:35






  • 2




    $begingroup$
    @Laura But a nice answer.
    $endgroup$
    – callculus
    Jan 13 at 22:47







1




1




$begingroup$
Thanks @callculus for pointing that out. Corrected!
$endgroup$
– pendermath
Jan 13 at 22:24




$begingroup$
Thanks @callculus for pointing that out. Corrected!
$endgroup$
– pendermath
Jan 13 at 22:24




1




1




$begingroup$
Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
$endgroup$
– pendermath
Jan 13 at 22:25




$begingroup$
Is it or is it not? :-) @callculus can you give more details? What's wrong with the solution?
$endgroup$
– pendermath
Jan 13 at 22:25




1




1




$begingroup$
@pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
$endgroup$
– callculus
Jan 13 at 22:26





$begingroup$
@pendermath It is the right answer. That´s why I´ve voted for it and I love it. I´ve just replied to Diogo Bastos comment.
$endgroup$
– callculus
Jan 13 at 22:26





2




2




$begingroup$
The solution of question 1941655 is also $9/13$. Am I missing something?
$endgroup$
– pendermath
Jan 13 at 22:35




$begingroup$
The solution of question 1941655 is also $9/13$. Am I missing something?
$endgroup$
– pendermath
Jan 13 at 22:35




2




2




$begingroup$
@Laura But a nice answer.
$endgroup$
– callculus
Jan 13 at 22:47




$begingroup$
@Laura But a nice answer.
$endgroup$
– callculus
Jan 13 at 22:47











3












$begingroup$

You can also do this with an example. Suppose there are $100$ employees. Then there are $60$ men, $0.6cdot60=36$ of whom are married, and there are $40$ women, $0.4cdot40=16$ of whom are married. There are then $36+16=52$ married workers, $36$ of whom are men, so the probability in question is $36over52=9over13$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Sure, it's just that @Laura wanted it solved using Bayes' Theorem
    $endgroup$
    – pendermath
    Jan 13 at 22:30















3












$begingroup$

You can also do this with an example. Suppose there are $100$ employees. Then there are $60$ men, $0.6cdot60=36$ of whom are married, and there are $40$ women, $0.4cdot40=16$ of whom are married. There are then $36+16=52$ married workers, $36$ of whom are men, so the probability in question is $36over52=9over13$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Sure, it's just that @Laura wanted it solved using Bayes' Theorem
    $endgroup$
    – pendermath
    Jan 13 at 22:30













3












3








3





$begingroup$

You can also do this with an example. Suppose there are $100$ employees. Then there are $60$ men, $0.6cdot60=36$ of whom are married, and there are $40$ women, $0.4cdot40=16$ of whom are married. There are then $36+16=52$ married workers, $36$ of whom are men, so the probability in question is $36over52=9over13$.






share|cite|improve this answer









$endgroup$



You can also do this with an example. Suppose there are $100$ employees. Then there are $60$ men, $0.6cdot60=36$ of whom are married, and there are $40$ women, $0.4cdot40=16$ of whom are married. There are then $36+16=52$ married workers, $36$ of whom are men, so the probability in question is $36over52=9over13$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 22:27









Steve KassSteve Kass

11.2k11430




11.2k11430











  • $begingroup$
    Sure, it's just that @Laura wanted it solved using Bayes' Theorem
    $endgroup$
    – pendermath
    Jan 13 at 22:30
















  • $begingroup$
    Sure, it's just that @Laura wanted it solved using Bayes' Theorem
    $endgroup$
    – pendermath
    Jan 13 at 22:30















$begingroup$
Sure, it's just that @Laura wanted it solved using Bayes' Theorem
$endgroup$
– pendermath
Jan 13 at 22:30




$begingroup$
Sure, it's just that @Laura wanted it solved using Bayes' Theorem
$endgroup$
– pendermath
Jan 13 at 22:30

















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