Alternative way to solve complement probability?

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Four people are chosen at random. What is the probability that:

No two of them have their birthday in the same month?

The first and second people chosen have their birthdays in the same month, given that there is some pair of people having their birthdays in the same month?

For the first question I think it's straight forward enough, P(no two of them have their birthday in the same month) = P(Everyone's birthday is the same) = $frac 1212 *frac 1112*frac 1012*frac 912$

For the second question I used Bayes' Theorem the set up I have is:

$P(1st ,2nd ,Same|P(Some, pair, same) = frac P(some , pair, same,)P(Some, pair, same) = frac 1* (frac1212*frac112)binom42 frac1212 frac112 frac1112 frac1112$

The solution seems to suggest that P(Some pair having the same birthday) = $1- frac 1212 *frac 1112*frac 1012*frac 912$ which doesn't make any sense to me.

How can the complement of P(all birthdays are different) = P(some pair have same birthday). I thought the probability of some pair have the same birthday is in format like: AABC, ABCC, etc. Shouldn't the complement of P(all birthdays are different) means that Not all birthdays are the same meaning that you could have cases like AAAB, ABBB ?

probability birthday

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edited Jul 14 at 22:02

Henry

93.1k469147

asked Jul 14 at 21:37

pino231

3268

add a comment | 

up vote
0
down vote

favorite

Four people are chosen at random. What is the probability that:

No two of them have their birthday in the same month?

The first and second people chosen have their birthdays in the same month, given that there is some pair of people having their birthdays in the same month?

For the first question I think it's straight forward enough, P(no two of them have their birthday in the same month) = P(Everyone's birthday is the same) = $frac 1212 *frac 1112*frac 1012*frac 912$

For the second question I used Bayes' Theorem the set up I have is:

$P(1st ,2nd ,Same|P(Some, pair, same) = frac P(some , pair, same,)P(Some, pair, same) = frac 1* (frac1212*frac112)binom42 frac1212 frac112 frac1112 frac1112$

The solution seems to suggest that P(Some pair having the same birthday) = $1- frac 1212 *frac 1112*frac 1012*frac 912$ which doesn't make any sense to me.

How can the complement of P(all birthdays are different) = P(some pair have same birthday). I thought the probability of some pair have the same birthday is in format like: AABC, ABCC, etc. Shouldn't the complement of P(all birthdays are different) means that Not all birthdays are the same meaning that you could have cases like AAAB, ABBB ?

probability birthday

share|cite|improve this question

edited Jul 14 at 22:02

Henry

93.1k469147

asked Jul 14 at 21:37

pino231

3268

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Four people are chosen at random. What is the probability that:

No two of them have their birthday in the same month?

The first and second people chosen have their birthdays in the same month, given that there is some pair of people having their birthdays in the same month?

For the first question I think it's straight forward enough, P(no two of them have their birthday in the same month) = P(Everyone's birthday is the same) = $frac 1212 *frac 1112*frac 1012*frac 912$

For the second question I used Bayes' Theorem the set up I have is:

$P(1st ,2nd ,Same|P(Some, pair, same) = frac P(some , pair, same,)P(Some, pair, same) = frac 1* (frac1212*frac112)binom42 frac1212 frac112 frac1112 frac1112$

The solution seems to suggest that P(Some pair having the same birthday) = $1- frac 1212 *frac 1112*frac 1012*frac 912$ which doesn't make any sense to me.

How can the complement of P(all birthdays are different) = P(some pair have same birthday). I thought the probability of some pair have the same birthday is in format like: AABC, ABCC, etc. Shouldn't the complement of P(all birthdays are different) means that Not all birthdays are the same meaning that you could have cases like AAAB, ABBB ?

probability birthday

share|cite|improve this question

edited Jul 14 at 22:02

Henry

93.1k469147

asked Jul 14 at 21:37

pino231

3268

Four people are chosen at random. What is the probability that:

No two of them have their birthday in the same month?

The first and second people chosen have their birthdays in the same month, given that there is some pair of people having their birthdays in the same month?

For the first question I think it's straight forward enough, P(no two of them have their birthday in the same month) = P(Everyone's birthday is the same) = $frac 1212 *frac 1112*frac 1012*frac 912$

For the second question I used Bayes' Theorem the set up I have is:

$P(1st ,2nd ,Same|P(Some, pair, same) = frac P(some , pair, same,)P(Some, pair, same) = frac 1* (frac1212*frac112)binom42 frac1212 frac112 frac1112 frac1112$

The solution seems to suggest that P(Some pair having the same birthday) = $1- frac 1212 *frac 1112*frac 1012*frac 912$ which doesn't make any sense to me.

How can the complement of P(all birthdays are different) = P(some pair have same birthday). I thought the probability of some pair have the same birthday is in format like: AABC, ABCC, etc. Shouldn't the complement of P(all birthdays are different) means that Not all birthdays are the same meaning that you could have cases like AAAB, ABBB ?

probability birthday

share|cite|improve this question

edited Jul 14 at 22:02

Henry

93.1k469147

asked Jul 14 at 21:37

pino231

3268

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 22:02

Henry

93.1k469147

edited Jul 14 at 22:02

Henry

93.1k469147

edited Jul 14 at 22:02

Henry

93.1k469147

93.1k469147

asked Jul 14 at 21:37

pino231

3268

asked Jul 14 at 21:37

pino231

3268

asked Jul 14 at 21:37

pino231

3268

3268

add a comment | 

add a comment | 

1 Answer
1

active

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votes

up vote
1
down vote



accepted

It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs

If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $dfrac14 choose 2$

But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $dfracfrac1212times frac112 1- frac 1212 timesfrac 1112timesfrac 1012timesfrac 912$

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
  – pino231
  Jul 14 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
  – Henry
  Jul 15 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
  – pino231
  Jul 16 at 1:05
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs

If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $dfrac14 choose 2$

But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $dfracfrac1212times frac112 1- frac 1212 timesfrac 1112timesfrac 1012timesfrac 912$

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
  – pino231
  Jul 14 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
  – Henry
  Jul 15 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
  – pino231
  Jul 16 at 1:05
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs

If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $dfrac14 choose 2$

But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $dfracfrac1212times frac112 1- frac 1212 timesfrac 1112timesfrac 1012timesfrac 912$

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
  – pino231
  Jul 14 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
  – Henry
  Jul 15 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
  – pino231
  Jul 16 at 1:05
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs

If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $dfrac14 choose 2$

But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $dfracfrac1212times frac112 1- frac 1212 timesfrac 1112timesfrac 1012timesfrac 912$

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs

If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $dfrac14 choose 2$

But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $dfracfrac1212times frac112 1- frac 1212 timesfrac 1112timesfrac 1012timesfrac 912$

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 22:02

Henry

93.1k469147

answered Jul 14 at 22:02

Henry

93.1k469147

answered Jul 14 at 22:02

Henry

93.1k469147

93.1k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
  – pino231
  Jul 14 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
  – Henry
  Jul 15 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
  – pino231
  Jul 16 at 1:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
  – pino231
  Jul 14 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
  – Henry
  Jul 15 at 11:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
  – pino231
  Jul 16 at 1:05
  

  
  
  
  

Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
– pino231
Jul 14 at 23:58

Regarding the scenario The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? Shouldn't the answer be $frac 1* frac112binom42 frac1212 frac112 frac1112 frac1112$ ? Shouldn't the P(Exactly 1 pair)= $binom42 frac1212 frac112 frac1112 frac1112$
– pino231
Jul 14 at 23:58

1

1

@Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
– Henry
Jul 15 at 11:06

@Pino: close but not quite: the probability of exactly one pair is $binom42 frac1212 frac112 frac1112 frac1012$
– Henry
Jul 15 at 11:06

oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
– pino231
Jul 16 at 1:05

oh yea you're right I don't know what I was thinking, too transfixed on the fist two terms. Thanks!
– pino231
Jul 16 at 1:05

add a comment | 

 

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