English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Consider the population of all families with two children. Representing the gender of each child using G for girl and B for boy results in four possibilities: BB, BG, GB, and GG. The gender information is sequential, with the first letter indicating the gender of the older sibling. Thus, a family having a girl first and then a boy would be denoted by a GB.

If we assume that a child is equally likely to be male or female, each of the four possibilities in the sample space for the experiment that selects at random from families with two children is equally likely.

2007-12-26 15:17:42 · 6 answers · asked by Jerry 2 in Education & Reference Homework Help

6 answers

I believe that it would be 25%, right? There are four possible combinations, each as likely as the next, considering there is a 50/50 chance of getting a boy or a girl. So each combination is 25% likely to be selected in a random selection from families that definitely have two children.

---OHH okay, I didn't understand the question at first; I thought this was about random selection. In which case my answer was wrong anyway, because if it already had one girl, then the chance that the other kid is a girl is 33%.
If the question is "What's the chance the next kid is a girl?", its the same no matter what you already have, even if you have ten girls, its still 50/50.

2007-12-26 15:46:08 · answer #1 · answered by Zoe - Little Linguist 4 · 0 3

This is an exercise in conditional probability

For any two events A and B, where P(B) ≠ 0, you have the conditional probability:

P( A | B ) = P( A ∩ B ) / P( B )

the above is read as: the probability of A given B is equal to the probability of A and B divided by the probability of B.

Let A be the event of having two girls
Let B be the event of having at least one girl

P(A) = P(GG) = 1/4
P(B) = P(BG or GB or GG) = 3/4

P( A | B ) = P(A and B) / P(B)

P(A and B) = P(GG) = P(A) = 1/4

P(A | B) = (1/4) / (3/4)
P(A | B) = 1/3

Given that the family has at least one girl, the probability they have two girls is 1/3.

2007-12-27 02:58:46 · answer #2 · answered by Merlyn 7 · 0 0

Since BB features no girls, the possibilities are BG, GB and GG - all equally likely.

So the probability of two girls is 1/3.

2007-12-26 16:30:50 · answer #3 · answered by Beardo 7 · 0 1

50%, though there is only a 25% chance of a couple with no children having 2 girls, a 50% chance of them having a boy and a girl, and a 25 % chance of them having 2 boys. Once the couple has the 1st girl, there can only be a chance of having a boy or a girl, there fore a 50% chance of having a girl once the couple already has a girl

2007-12-26 15:49:55 · answer #4 · answered by ? 3 · 0 0

I was told that the sex of the first child was most likely to be the sex of the second child in two out of three cases. So, if you have a boy you are more likely to have a second boy. Only one out of three will have a child of each sex.
Don't know how accurate this is but I did observe these results in a small group of expectant couples.

2007-12-26 15:49:45 · answer #5 · answered by Anonymous · 0 1

One half. Fifty percent. One out of two.

Since you have a girl already, the possibilities for a second child are GB and GG. Of the TWO possibilities, ONE of those possibilities satisfies our query. One out of two.

2007-12-26 15:46:10 · answer #6 · answered by Sage B 4 · 2 0

fedest.com, questions and answers