I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?
Start with an apparently simpler problem:
"I have two children. One is a boy. What is the probability I have two boys?"
It might be more obvious that 50% is the wrong answer to that one.
Well if you didn't tell me that one is a boy. I would say the chances are 1/4 since there are BB BG GB GG. But given that you have told me one is a boy...I'm either thinking 1/2 still or 1/3.
I'm going to go with 1/3 for this one. If the question were "I have two children. The first is a boy. What is the probability that I have two boys?", then the answer would be 50%.
None of the numbers in this thread are answers to the original question yet.
Is it perhaps 6/14? (I'm only guessing this based on the fact that the 1/2 is wrong, but I still feel it should be 1/2 :tongue:)
I won't explain myself unless I am confirmed correct (for the sake of not giving out too many hints).
In order to be another boy it NEEDS to be born on another Tuesday. There are 7 days a week and you can only have another every 9 months. Once the ninth month is hit there are only 21 days available for the child to be born. Not more, not less. In those 21 there are only 3 Tuesdays, which means:
3/21 = 1/7 = 14.something%
But all the other days are girls, which means there's somewhat 85% chance it will be a girl. Therefore, you had roughly 14% chances of having another boy.
Do I need to know this?
Is the answer slightly under 1/3 because of the probability of a hermaphrodite child?
No. To clarify, here are the rules:
1) we’re ruling out twins,
2) we’re pretending that boys and girls are born with equal likelihood, and
3) we’re assuming that there is no correlation between a child’s sex and the day of the week on which they’re born.
Well, I guess we can list out all possibilities. The first column being the gender and the second column being the day:
BMBM BMBT BMBW BMBTH BMBF BMBS BMBSU
BTBM BTBT BTBW BTBTH BTBF BTBS BTBSU ...
BMGM BMGT BMGW BMGTH BMGF BMGS BMBSU...
GMBM GMBT GMBW GMBTH GMBF GMBS GMBSU...
GMGM GMGT GMGW GMGTH GMGF GMGS GMGSU...
Each block has 49 entries. The GG block is eliminated.
In the first block there are 13 entries with at least 1 boy born on Tuesday. In the second block there are 7 entries with a boy born on Tuesday, and in the third block there are 7 entries with a boy born on Tuesday. That's 27 total entries with a boy born at least on Tuesday. Of those 27 entries 13 are both boys while the other 14 have girls in them. So the probability is 13/27.
1st child Tuesday boy, 2nd girl on random day=7
1st child Tuesday boy, 2nd boy on random day=7
2nd child Tuesday boy, 1st girl on random day=7
2nd child Tuesday boy, 1st boy on random day=7
EDIT: nvm, too slow...
Not enough information.
You never said you have exactly two children.
Well, you don't need to know it was Tuesday. The probability would be the same for any named day of the week.
Of course that logic suggests another way to get the wrong answer: find the probability of some other situation, and divide it by 7
Here's how I'd look at it:
Given you have exactly two children, imagine asking each family with exactly two children the same question. Do you have a boy born on a Tuesday?
Typical families are: GG (25%), GB (50%), BB (25%).
The families with two girls always answer no, and the families with two boys are more likely to answer yes (with two boys, it's more likely one was born on a Tuesday). How much more likely depends on the rarity of the required property (for a rare property, it would be almost double; for a common property it may make very little difference).
In this case, it's almost double (roughly) so, the GB and BB cases are almost equally likely.
However, if you have simply volunteered this information without my asking, I'm not so sure:
You go to every family and they are asked to say whether they have a boy and, if so, volunteer a day of the week on which the boy was born, then you might not reach the same conclusion for every family with at least one boy:
The first family might say: we have a boy born on Monday. The second family might say: we have a boy born on Thursday. The third family might say: we have no boys. The fourth family might say: we have a boy born on Tuesday.
In this case, the probability of 2 boys is 1/3 as the additional information in each case implies nothing.
PeroK starts in the right place.
There are three possibilities per child, which I will label as G, for a girl, T for a boy born on Tuesday, and N for a boy not born on a Tuesday. Ordering the children, here are the outcomes and their probabilities:
GG = 1/4 = 49/196
GN = 3/14 = 42/196
NG = 3/14 = 42/196
GT = 1/28 = 7/196 *
TG = 1/28 = 7/196 *
NN = 9/49 = 36/196
NT = 3/98 = 6/196 *
TN = 3/98 = 6/196 *
TT = 1/196 = 1/196 *
We are told one child is a boy born on Tuesday. That means we are in one of the 5 asterisked outcomes.
Of them, the outcomes with two boys are NT, TN and TT, for a probability of 13/27.
I believe it makes a big difference if the information is volunteered. All families with at least one boy can quote a day of the week on which a son was born. This volunteered information does not exclude any family nor make it more likely they have two boys.
On the other hand if the family is asked whether they have a boy born on Tuesday, this does exclude many families and excludes more who have only one boy.
These are two different situations.
By volunteering this information the problem does not work as intended. The day of the week must be chosen by an external questioner and not chosen by the family.
Every family with at least one boy could do what Micomass did and say: We have a boy born on xxxday.
Separate names with a comma.