Probability?

Question

Find the probability of a couple having a baby girl when their third child is born, given that the first two children were both girls. is the results the same as the probability of getting three girls among three children?

Answer 1

50%

The gender of one child is independent of the gender of its siblings.

Therefore it doesn't matter if you had 6 girls prior, or 8 boys, or whatever, the probability that THIS child will be a girl is 50%.

(Similar to how if you toss a coin, you always have a 50% chance of heads, regardless of how many heads or tails you've thrown beforehand.)

The genders of the previous children are thrown into the problem as a trick. Your job in solving the problem is to realize that the genders of the babies are independent events that have no bearing on the matter at hand.

To answer the second part of your question: The probability of having 3 girls out of 3 children is .50 * .50 * .50 = .125. But that is totally different from the first part of your question (hopefully that's clear; if not feel free to ask another question).

~ ♥ ~

Answer 2

The probability of getting 3 girls among 3 children is (1/2)^3 = 1/8, so it is not the same as having a girl when the 3rd child is born, given the first two were girls. Each birth is an independent event, so the fact that the first two were girls has no bearing on the probability of the 3rd being a girl. So, the first question examines only a single event, while the second questin examines 3.

Answer 3

The answer to the first part is 50%. Just like flipping a coin thrice, the first two results have zero bearing on the third flip.

The results are not the same as the question of getting all three girls, which would be considerably less than 50%.

Answer 4

I suspect my argument isn't going to be good got. I say the likelihood is 50% Let a ??, Let b ?? and randomly prefer the values for a and b. As already famous, for a ? zero, P( a < b²) = one million, that is trivial. Only rather much less trivial is the proposal that P(a < zero ) = one million/two and accordingly P( a < b² | a ? zero) = one million and P( a < b² ) ? one million/two Now don't forget what occurs while a > zero For a > zero, even as it's convenient to exhibit there's a non 0 likelihood for a finite b, the restrict, the likelihood is 0. a < b² is identical to pronouncing zero < a < b², bear in mind we're most effective watching at a > zero. If this a finite period on a vast line. The likelihood that a is an detail of this period is 0. P( a < b² | a > zero) = zero As such now we have a whole likelihood P( a < b² ) = P( a < b² | a ? zero) * P(a ? zero) + P( a < b² | a > zero) * P(a > zero) = one million * one million/two + zero * one million/two = one million/two Remember, that is due to the fact that of the limitless units. No subject what variety of period you draw on paper or on a laptop you'll discover a finite likelihood that looks to system one million. But that is because of the finite random quantity mills at the laptop and if we had this question requested with finite values there could be a an answer higher than 50%. I do not imply to be condescending, however please provide an explanation for why utilizing the Gaussian to approximate a uniform distribution is a well proposal? Aren't limitless numbers amusing. Cantor while mad operating with them! :)