# Two Girls A Boy Sex

As the calculation is already shown in another answer, I won't repeat it here, only give the rough answers:P(A)=P(all boys) #1/8#P(B)=P(all boys or all girls)#1/4#P(C)=P(2 boys or 2 girls)#3/4#P(D)=P(at least 1 of each gender)#3/4#

## two girls a boy sex

Perhaps the best-known reason relates to the practice of sex-selective abortion, which has been identified in Asia, and in the Caucasus, as well. The ability to determine fetal sex, along with strong son preferences, accounts in large part for the high shares of boys in many countries in these regions. The desire to limit family size, either due to government regulations as in China, or due to global social and economic changes that have reduced the need for large families, seems to further contribute to sex-selective abortion and a dearth of baby girls.

But this is only one of myriad factors that may be affecting the sex ratio at birth. Some research suggests that the share of newborn boys declines with older parents, and that the high share of girls in Sub-Saharan Africa may be linked to the practice of polygamy (multiple wives). What do these two phenomena have in common? Researchers hypothesize that both situations are associated with less frequent intercourse. (For possible explanations of this association, see this article from the academic journal Human Reproduction.)

The two possible answers share a number of assumptions. First, it is assumed that the space of all possible events can be easily enumerated, providing an extensional definition of outcomes: BB, BG, GB, GG.[10] This notation indicates that there are four possible combinations of children, labeling boys B and girls G, and using the first letter to represent the older child. Second, it is assumed that these outcomes are equally probable.[10] This implies the following model, a Bernoulli process with p = 1/2:

Only two of these possible events meet the criteria specified in the question (i.e., GG, GB). Since both of the two possibilities in the new sample space GG, GB are equally likely, and only one of the two, GG, includes two girls, the probability that the younger child is also a girl is 1/2.

For example, say an observer sees Mr. Smith on a walk with just one of his children. If he has two boys then that child must be a boy. But if he has a boy and a girl, that child could have been a girl. So seeing him with a boy eliminates not only the combinations where he has two girls, but also the combinations where he has a son and a daughter and chooses the daughter to walk with.

Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is 1/2 and, when making the most natural assumptions, this is correct. However, someone may argue that "...before Mr. Smith identifies the boy as his son, we know only that he is either the father of two boys, BB, or of two girls, GG, or of one of each in either birth order, i.e., BG or GB. Assuming again independence and equiprobability, we begin with a probability of 1/4 that Smith is the father of two boys. Discovering that he has at least one boy rules out the event GG. Since the remaining three events were equiprobable, we obtain a probability of 1/3 for BB."[3]

The natural assumption is that Mr. Smith selected the child companion at random. If so, as combination BB has twice the probability of either BG or GB of having resulted in the boy walking companion (and combination GG has zero probability, ruling it out), the union of events BG and GB becomes equiprobable with event BB, and so the chance that the other child is also a boy is 1/2. Bar-Hillel & Falk, however, suggest an alternative scenario. They imagine a culture in which boys are invariably chosen over girls as walking companions. In this case, the combinations of BB, BG and GB are assumed equally likely to have resulted in the boy walking companion, and thus the probability that the other child is also a boy is 1/3.

However, the exact sex ratio at birth for males to females is approximately 105:100. This is a little more than expected. Explanations include a strong preference for boys worldwide which results in about 130 million missing girls due to selective abortion, infanticide, or neglect.

Although it is the man's sperm that ultimately determines the sex of a baby, is it a coincidence that some women give birth to only boys while others only have baby girls? Math and science have a lot to say about it!

One study suggests that looking at family trees could give some insight into why some couples keep having only girls or only boys. A Newcastle University study looked at the family history of nearly 1,000 couples to try to get down to the bottom of why some families are all girls or all boys. Researchers found that men are more likely to have sons if they have more brothers and are more likely to have daughters if they have sisters. But, in women, the likelihood of having a girl or boy just couldn't be predicted.

As we can tell from a variety of studies, more research needs to be done into why women only give birth to boys or girls. Much of the research has been focused on men because they have the deciding chromosome, but more studies can be done to look at womens' family trees to see if any type of pattern can be determined.

Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac13$.

Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.

On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.

Do you see the difference? In the first case, every family has the same probability of being chosen. However, in the second case, the families with two girls are more likely to be chosen than families with only one girl, because every girl has an equal chance of being chosen: the two-girl families have doubled their odds by having two girls. If children were raffle tickets, they would have bought two tickets while the one-girl families bought only one.

Suppose you sample $4n$ families with two children, where $n$ is a very large integer. Then, with a very high probability, about $n$ of them have two boys, about $n$ have two girls, and about $2n$ have one girl and one boy.Since you ignore the families with two boys, the desired probability is given by $n/(3n) = 1/3$.

EDIT: The above solution corresponds to the following situation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you check if one of the kids is a girl. If no, you ignore this family. If yes, you check if both kids are girls.

EDIT: Among all the OP's examples, only one is relevant here. Namely, "I visit this family. I know they have 2 kids. One of them, a girl, comes into the room. The probability that the 2nd kid is also a girl is 1/2, no?". Well, in this situation the probability is indeed $1/2$, assuming the following interpretation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you wait until one of the kids comes into the room. If it is a boy, you ignore this family; otherwise you check if both kids are girls. It is readily understood that this leads to a probability of $1/2$.

You have to read carefully. Here, the probability of a girl means the probability of at least one girl. You have$$P(\hboxone child is a girl) = 1 - P(\hboxno child is a girl)= 1 - P(\hboxtwo boys) = \frac14. $$Now you are calculating the conditional probabilty$$P(\hboxtwo girls \hboxat least 1 girl = P(\hboxtwo girls)\over P(\hboxat least one girl) = 1\over 3.$$

However, if it is known that at least one of the children is a girl, then the event represented by the bottom branch can't happen. Of the remaining three equiprobable events, only one corresponds to the event that there are two girls. Therefore if it is known that a family has two children and it is further known that one of those children is a girl, then the probability that the family has two girls is $1/3$.

The problem asks the probability of having both girls if you were told one of the children is a girl. The trick of this problem is that you still do not know this girl is from the first child or the 2nd child. Thus, this shows us a new sample space$\Omega'=\gg,gb,bg\$. Then the probability of having both girls is just 1/3.

More interesting thing is to study where 1/2 comes from. I present a scenario where the answer is 1/2. If you were told that the first child is a girl, the sample space becomes$\Omega'=\gg,gb\$. In this case, the probability of having both girls is 1/2.

This is shown in Figure 2-6. It is very rare for a man with hemophilia and a woman who is a carrier to get together. That is why there have been only a few girls born with hemophilia.Figure 2-6. Father with hemophilia; mother a carrier. Follow the arrows to see the possible gene combinations.

The purpose was to discover whether the preschool child who has toileted with boys and girls in an open situation for at least four weeks would recognize the genital differences between a male and a female doll and use this knowledge to select the type of clothes appropriate for these dolls. Hypotheses were: (1) The preschool child who has toileted in an open situation with both sexes for at least four weeks can identify male and female dolls on the basis of genital structure. (2) The preschool child can select appropriate clothes for male and female dolls on the basis of genital structure as the dominant cue as to sexuality . Twenty children ages three to five were selected because they had toileted in an open situation for at least four weeks with boys and girls, because they attended the same classroom, and were familiar with the experimenter . Preceding the actual collection of data a pilot study was conducted on a similar but separate classroom of children to test the proposed research design. The subjects were informed, as a group, of the general content and procedure of the experimental situation to follow. The subjects went first behind the screen with the experimenter to arrange six blocks in any design they wished. They went again behind the screen in the doll house where they sat facing the genitals of a male and a female doll. After powdering the dolls' genitals, they selected clothes for them from two sets of male slacks and two dresses. The subject was then asked the reason for his choice. Four of the 20 subjects, two boys and two girls, were able to identify the sex of the dolls on the basis of genital structure. All four were from homes with opposite-sex siblings. Three of the four expressing a recognition were four years old . Conclusions: (l) Preschool children appear generally to be unable to identify sexuality on the basis of genital structure . (2) Age is a significant factor influencing the child's discrimination of sexual differences. (3) The opportunity for observation and discovery of differences between the sexes through the presence of opposite-sex siblings in the home appears to be influential on the child's development of sexual awareness.