I dont get how its 1/3. My logic: It says we already have 1 girl, so we can ignore her and just look at the other one, thus answer: 1/2.
Look at the sample space
You have {(B,B), (B,G), (G,G)}
The question asks you for the odds of having (G,G), which happens to be \frac 13 .
Edit: Typo of (B,B) instead of (B,G)
But (B,B) is impossible since it is given that we already have 1 girl.
The sample space is the set of all possible outcomes.
So you can have any 2-combination of girls and boys, which was what i listed out. There’s \frac 13 chance of picking (G,G) because all of the elements in the sample space have equal probability.
No, you are wrong. If it is already given that one of them is a girl the case of both being boys is impossible and thus we don’t include it in the sample space.
I’m not wrong about the sample space, but i will admit i skipped a bit of details, so here’s a more thorough explanation.
Anyhow, the sample space is all 2-element permutations of boy/girls, and has nothing to do with what you want. You choose what u want from the sample space and assign it a probability.
In this case, the sample space is:
\Omega = \{ (B,B), (G,G), (B,G), (G,B)\}
All the elements in this sample space are equiprobable.
Now this is a conditional probability exercise:
Given that you have a girl ( no position is mentioned so any of these work):
(B,G), (G,B), (G,G)
What are the odds of picking (G,G)?
As for your initial logic, we don’t know anything about the position of the girl, so we have to make cases for it.
In the process, make sure to not over count (G,G) cuz that’ll lead you to the wrong answer of \frac 12.
You are absolutely right, the answer is 1/2. This is the second forum post about a mistake in a conditional probability question I see today, curious. Also, cylverixxx is wrong about the sample space. The reason is that we can order the elements by identifying the first entry with the child we already know is a girl and the second being the one we don’t know. The correct sample space is then {(G,G), (G,B)}. The event (B,G) is not in the sample space because that would correspond to the impossible event that the child we already knew was a girl is actually a boy.
EDIT: By the way, this way of thinking about orderings is very confusing, and that’s why you should just use the property P(A | B) = P(A) if A and B are independent (notation: P(A | B) means “the probability of A conditional on B”).
Cylverixxx, imagine the question was about a couple with a thousand children, and we know they have 999 girls. What is the probability that all 1000 children are girls?
No, where does it say the first child must be a girl? If the first child is fixed to be a girl, then you definitely don’t need conditional probability to argue that the answer is \frac 12
It does not say that. But what is the first child? First to be born? First name alphabetical order? Most loved by mom? No, its however we can order them, and we can order them by the one we know and the one we don’t. Again, this is confusing and certainly not the best way to approach the question. the combinatorics of conditional probability is weird and its best to think in terms of probability theory, as explained in the edit. If you really want to approach it using combinatorics, here’s another way of thinking about this: order by date of birth. If we know that one child is a girl, we could be in one of two equally probable worlds: either the first is a girl {(G,G), (G, B)} or the second is a girl: {(G,G), (B, G)}. In either case, the probability of the other child being a girl is 1/2, which can be verified using combinatorics. So the answer is 1/2 * 1/2 + 1/2 * 1/2 = 1/2. In english: the probability of being in world A times the probability that the second child is a girl plus the probability of being in world B times the probability that the first child is a girl.
I don’t know why you’ve turned this into a philosophical question with the “what is the first child” 
. This is practically why i listed out all 2 element tuples in my sample space \Omega.
As for your actual argument, that is obviously wrong because you’re overcounting (G,G).
Let O be the event that the first is a girl
O=\{(G,B),(G,G)\}
Similarly, the event that the second is a girl can be represented as:
T=\{(B,G),(G,G)\}
The event that at least one is a girl is the union of O and T (O or T).
In fact, the union would be:
O \cup T = \{(B,G), (G,B), (G,G)\}
Just from this you can already see the probability of (G,G) is 1/3 lol
We also know that
P(O \cup T) = P(T) + P(O) only if P(O \cap T)=0.
(priori sets can’t have duplicate elements so we can’t count (G,G) twice)
Since O and T have (G,G) in common your formula doesn’t work and your reasoning is incorrect.
Man, I’m sorry but I don’t have the time. I seriously advise you to continue looking into this, because you will realize that I am right eventually. Ask a professor.
Gui, you keep saying “you’re right” whenever i point out flaws in your argument, but isn’t that confidence in your answer unfounded? I’m 100% sure I’m right, but i don’t keep saying that because it is irrelevant to the discussion.
For reference, here’s the same problem from a probability/stats book, but generalized for n children.
Now if you plug in n = 2, we get 1/3 (what a surprise!). I won’t get into the details anymore because you’re just being ignorant and whatever effort I put into this seems to go to waste.
Now if you plug in n = 2, we get 1/3 (what a surprise!)
Do you, tho?
Yes you do lol. I’m not sure if that’s supposed to be sarcasm of sorts. The solution presented is a “closed form” solution. In other words, you can obtain an expression for any value of n. The “n” happens to be the number of children, which is 2 in this case.
No, you don’t.
\frac 1 {2^{2-1}} = \frac 1 2
Too late man. You were snarky with me, then provided two sources that agree with me and embarrassed yourself by thinking they didn’t. I’m not being paid to teach you.
How does it feel to be wrong? For someone, that was constantly saying “I’m right”; it must’ve been really embarrassing. Also this is hilarious, you were being snarky to me throughout and implying all sorts of things; however, when i repeat the same thing on just a single message, you seem so affected.
Also you’re right, I did “embarrass” myself by posting a picture of the wrong question. I wasn’t paying much attention to it because I was a bit busy and it’s late at ng. If you’re interested why those pictures give different answers, it is because the family could have a boy and girl but just not have the girl be the one picked.
Effectively, for the “wrong question i linked”, for n = 2 the space is more like
bb → b
bb → b
bg → b
bg → g
gb → g
gb → b
gg → g
gg → g
there are 2 copies of ‘g’, and in half of them, both children are girls so yeah… clearly different questions lol.
If you had even a sliver of probability knowledge, you’d have pointed out that the questions are “different”.
Anyway, I’ve posted a stackexchange thread of the exact question, and now we’re both convinced that you’re wrong. You should study conditional probability or pray to the GRE gods that they don’t test you on this. Don’t worry, you don’t have to admit that you’re wrong because deep down you know you’ve messed up (at least from the posted link ;p).
constantly saying “I’m right”
I said it once.
If you’re interested why those pictures give different answers, it is because the family could have a boy and girl but just not have the girl be the one picked.
YOU ARE RIGHT! Yep, I read the post and this is me conceding that the answer is 1/3. This is Monty Hall in disguise, but the Monty Hall setup is less ambiguous. The wording of the question is bad, but 1/3 is still technically correct.
If you had even a sliver of probability knowledge, you’d have pointed out that the questions are “different”.
Seems like neither of us do. Or maybe you do, but you were busy and tired…
You should study conditional probability or pray to the GRE gods that they don’t test you on this.
Already got the 170 in quant. Good luck on your test and thanks for the interesting discussion!