Calculating Probability using Bayes Theorem
Clash Royale CLAN TAG#URR8PPP
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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
asked Jan 13 at 22:08
LauraLaura
2288
$endgroup$
add a comment |
$begingroup$
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
asked Jan 13 at 22:08
LauraLaura
2288
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 13 at 22:08
LauraLaura
2288
$endgroup$
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
asked Jan 13 at 22:08
LauraLaura
2288
asked Jan 13 at 22:08
LauraLaura
2288
asked Jan 13 at 22:08
LauraLaura
2288
asked Jan 13 at 22:08
LauraLaura
2288
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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!
answered Jan 13 at 22:19
pendermathpendermath
54410
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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
|
show 6 more comments
$begingroup$
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 $36over52=9over13$.
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
$endgroup$
$begingroup$
Sure, it's just that @Laura wanted it solved using Bayes' Theorem
$endgroup$
– pendermath
Jan 13 at 22:30
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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!
answered Jan 13 at 22:19
pendermathpendermath
54410
$endgroup$
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
|
show 6 more comments
$begingroup$
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!
answered Jan 13 at 22:19
pendermathpendermath
54410
$endgroup$
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
|
show 6 more comments
$begingroup$
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!
answered Jan 13 at 22:19
pendermathpendermath
54410
$endgroup$
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!
answered Jan 13 at 22:19
pendermathpendermath
54410
edited Jan 13 at 22:22
answered Jan 13 at 22:19
pendermathpendermath
54410
answered Jan 13 at 22:19
pendermathpendermath
54410
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
|
show 6 more comments
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
|
show 6 more comments
$begingroup$
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 $36over52=9over13$.
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
$endgroup$
$begingroup$
Sure, it's just that @Laura wanted it solved using Bayes' Theorem
$endgroup$
– pendermath
Jan 13 at 22:30
add a comment |
$begingroup$
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 $36over52=9over13$.
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
$endgroup$
$begingroup$
Sure, it's just that @Laura wanted it solved using Bayes' Theorem
$endgroup$
– pendermath
Jan 13 at 22:30
add a comment |
$begingroup$
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 $36over52=9over13$.
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
$endgroup$
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 $36over52=9over13$.
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
answered Jan 13 at 22:27
Steve KassSteve Kass
11.2k11430
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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