These problems seem so easy or am I just tricked by the wording?
1). You visit the home of an acquaintance, who says, "I have two kids." A boy walks into the room. The acquaintance says, "That's one of my kids." What is the probability that the other one is a boy?
2). You live in a culture where, when children are introduced, male children are always introduced first, in descending order of age, and then female children, also in descending age order. You visit the home of an acquaintance, who says, "I have two kids, let me introduce them." He yells, "John come here." (John is a boy's name). What is the probability that the other child is a boy?
3). You go to a parent-teacher meeting. The principal is sitting in the first row. You've heard that the principal has two children. The teacher in charge asks everyone who has a son (meaning at least one) to raise a hand. The principal raises her hand. What is the probability that the principal has two sons?
In the three problems, assume that a child is equally likely to be male or female and that the sex of one child in a family is independent of the sex of any other children in that family.
I'm getting really tricked by the wording, it seems to me that the answer to all of them is 1/2 but that's not the case for all three of them.
Nearly everyone is tricked by the wording. Those links mr fantastic provided included (even though I helped to write the wikipedia one, others wouldn't agree to the correct solutions). Your third problem requires a different solution than the first two, but people fail to understand why it is different, and sometimes apply the same solution where it is not appropriate.
1/2. The gender of the child you don't see is independent of the one you do.1). You visit the home of an acquaintance, who says, "I have two kids." A boy walks into the room. The acquaintance says, "That's one of my kids." What is the probability that the other one is a boy?
1/2. The gender of the younger child is independent of the older child's.2). You live in a culture where, when children are introduced, male children are always introduced first, in descending order of age, and then female children, also in descending age order. You visit the home of an acquaintance, who says, "I have two kids, let me introduce them." He yells, "John come here." (John is a boy's name). What is the probability that the other child is a boy?
1/3.3). You go to a parent-teacher meeting. The principal is sitting in the first row. You've heard that the principal has two children. The teacher in charge asks everyone who has a son (meaning at least one) to raise a hand. The principal raises her hand. What is the probability that the principal has two sons?
Working backwards:
3) From what you have heard, you know the the principal could equally likely have two boys (BB), and older boy and a younger girl (BG), an older girl and a younger boy (GB), or two girls (GG). Their ages are not directly important to your question, but they do allow us to treat the four cases as equally likely: each occurs 25% of the time.
But she could not have raised her hand if she had GG, so we can rule that case out. The three remaining cases are BB, BG, and GB. The probability of a BB familiy given that the family must be from BB, BG, or GB is found by dividing the probabilty of BB by the probability of BB, BG, or GB. So, P(BB)/(P(BB)+P(BG)+P(GB))=(25%)/(25%+25%+25%)=1/3.
The difference you are not seeing, is that the "boy" you learned about could be either the older, the younger, or actually both. It wasn't a specific child.
2) Here we can rule out both GB and GG. The probability of a BB family given that the family must be from BB, or BG is P(BB)/(P(BB)+P(BG))=(25%)/(25%+25%)=1/2.
1) People are tempted to use the solution for #3 here, but that is wrong. The boy you met can't be both the older and the younger.
There are several ways to do it correctly.
We can still eliminate the GG group. But the boy you met could be either the older or the younger, so we can't eliminate either BG or GB. But that doesn't mean we can count them the same way we did before. Half of the time with a BG or GB family, you would meet the girl this way. So each of those groups is given only half of its probability. The probability of a BB familiy given that the family must be from BB, half of BG, or half of GB is P(BB)/(P(BB)+P(BG)/2+P(GB)/2)=(25%)/(25%+12.5%+12.5%)=1/2.
-OR-
You can create eight groups, by including which child you would meet. The older, or the younger: BBO, BBY, BGO, BGY, GBO, GBY, GGO, and GGY. Each of these occure 12.5% of the time. You can eliminate those where you would meet a girl, leaving BBO, BBY, BGO, and GBY. The probability you want is (P(BBO)+P(BBY))/(P(BBO)+P(BBY)+P(BGO)+P(GBY))/2)=(12.5%+12.5)/(12.5%+12.5%+12.5%+12.5%)=1/2. You should be able to see how this relates to the first solution.
-OR-
As I said before, age is not important to your question, it just lets us create the groups. You could use any ordering to do the same thing, as long as it is random and independent of gender. You could do it by alphabetizing their names, using their height percentiles (which are adjusted by age and gender), or.... which child you met first. If that is used, then you know you met a boy first. The same solution as in #2 applies.