The Baby Nursery A nursery currently has 3 boys and x girls in it. Your wife if rushed to the hospital and gives birth to a baby. The baby is now added to the nursery. I go to the nursery and ask a nurse to randomly bring me a baby. She does, and it happens to be a baby boy. What is the probability that your wife gave birth to a boy?
It’s 4/7. For some reason, it doesn’t seem to matter what x is. Feel free to check my math, in case I made a mistake. 1) If we have no info, the P(B) = P(G) = 0.5 2) P(brings boy|B) = 4/(x+4) 3) P(brings boy|G) = 3/(x+4) Combine 1) and 2): P(brings boy, B) = 2/(x+4) Combine 1) and 3): P(brings boy, G) = 1.5/(x+4) Therefore, P(B|brings boy) = 2/(2+1.5) = 4/7
Anyone? Bueller? Bueller?
Hello Mister Walrus Wrote: ------------------------------------------------------- > Feel free to check my math, in case I made a > mistake. Looks good to me. Bonus points for explicitly stating reasonable simplifying assumption that P(B) = P(G). Interestingly, P(B) does not equal P(G) in the real world, and the result is statistically significant.
The better question is, “why are you going to the nursery to see my wife’s newborn?!?” Police!! (or divorce lawyer?!)
Nice - you got it. Yeah, number of girls doesn’t matter.
What? Probability of your wife giving birth to a boy is 50%, it doesn’t matter what child the nurse brings you. The two events are unrelated here and can’t be treated as a conditional probability.
cfagoal2 Wrote: ------------------------------------------------------- > What? Probability of your wife giving birth to a > boy is 50% Yes. > it doesn’t matter what child the nurse brings you. No. > The two events are unrelated here and can’t be > treated as a conditional probability. What if there were no other babies in the nursery?
All this assumes that the P(G) = P(B) = 0.5; which isn’t necessarily the case. Granted, its a reasonable assumption.
Interesting, I get 4/7 too, independent of the number of girls in the nursery. The weird thing is that if x=0, the chance that wifey delivered a boy is still 4/7. Not sure why that is, but these Bayes’ Theorem problems typically warp my mind. Yes, it’s true that, barring deliberate female-selected abortions, a higher proportion girls are conceived (x-chromasome sperm seem to live longer inside of a woman, though the y-chromasome sperm appear to swim faster), and a higher proportion of boys die in the womb. A higher proportion of boys also die in the first five years of life. I’ve heard a number of feminists claim that this “proves” that women are actually more biologically robust than men, though it could also be that people feel freer to neglect boys than girls, or that girls will call for help/attention more readily than boys. Men also die on average about 4 years earlier than women, but we knew this already.
bchadwick Wrote: ------------------------------------------------------- > The weird thing is that if x=0, the chance that > wifey delivered a boy is still 4/7. Why is that weird? > Yes, it’s true that, barring deliberate > female-selected abortions, a higher proportion > girls are conceived (x-chromasome sperm seem to > live longer inside of a woman, though the > y-chromasome sperm appear to swim faster), and a > higher proportion of boys die in the womb. These statements taken together are not consistent with the data we have on sex ratio at birth, where there is a clear male bias. There is a 1.05-1.07:1 ratio of male births to female births. (Look at CIA world factbook data on any first-world country like USA, Germany, Japan, etc.) This ratio levels out by early adulthood, which is what you would expect from an evolutionary perspective.
B - baby boy is born with unconditional probability P(B). A - boy is selected y boys (in this case 3), x girls. Looking for P(B|A). Notice, that P(A|B)=(y+1)/(y+x+1), P(A|not B)=y/(y+x+1)-> P(A)=(2*y+1)/(y+x+1). P(AB)=P(A|B)*P(B)=P(B|A)*P(A) -> P(B|A)=P(A|B)*P(B)/P(A)=P(B)*(y+1)/(2*y+1) since y = 3 -> P(B|A)=P(B)*4/7
Full-scale Bayesian Networks for complex problems are really fun, but you can easily get lost watching a wall-sized diagram, specially if they are dynamic.
Lets forget hte technicalities and assume the odds are 50-50. I don’t think which child is brought should impact the odds of what child you had. I could see them impacting the odds of whether a boy or girl was brought to you, but not in reverse.
It’s counterintuitive, but two people have already submitted proofs above.
maratikus Wrote: ------------------------------------------------------- > B - baby boy is born with unconditional > probability P(B). > A - boy is selected > > y boys (in this case 3), x girls. > > Looking for P(B|A). > > Notice, that P(A|B)=(y+1)/(y+x+1), P(A|not > B)=y/(y+x+1)-> P(A)=(2*y+1)/(y+x+1). > > P(AB)=P(A|B)*P(B)=P(B|A)*P(A) -> > P(B|A)=P(A|B)*P(B)/P(A)=P(B)*(y+1)/(2*y+1) > > since y = 3 -> P(B|A)=P(B)*4/7 P(A)=P(A|B)*P(B)+P(A|not B)*P(not B) i think you used P(A)=P(A|B)+P(A|not B) which gives you incorrect P(A) (for instance, it will be bigger than 1 for x
Mobius Striptease Wrote: ------------------------------------------------------- > maratikus Wrote: > -------------------------------------------------- > ----- > > B - baby boy is born with unconditional > > probability P(B). > > A - boy is selected > > > > y boys (in this case 3), x girls. > > > > Looking for P(B|A). > > > > Notice, that P(A|B)=(y+1)/(y+x+1), P(A|not > > B)=y/(y+x+1)-> P(A)=(2*y+1)/(y+x+1). > > > > P(AB)=P(A|B)*P(B)=P(B|A)*P(A) -> > > P(B|A)=P(A|B)*P(B)/P(A)=P(B)*(y+1)/(2*y+1) > > > > since y = 3 -> P(B|A)=P(B)*4/7 > > > P(A)=P(A|B)*P(B)+P(A|not B)*P(not B) > > i think you used P(A)=P(A|B)+P(A|not B) which > gives you incorrect P(A) (for instance, it will be > bigger than 1 for x
@Swan: it does because the wife’s child is added to the overall pool of children in the nursery.
Black Swan Wrote: ------------------------------------------------------- > Lets forget hte technicalities and assume the odds > are 50-50. I don’t think which child is brought > should impact the odds of what child you had. I > could see them impacting the odds of whether a boy > or girl was brought to you, but not in reverse. uhm, let’s say there were zero boys initially and x girls, and you know this. they brought you a boy. given that info, what is the probability that your wife gave birth to a boy? obviously it’s 1 and not 50-50
4/7