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        riddles >> easy >> simple conditional probability qns
        
(Message started by: Valerfore on Oct 29^th, 2004, 6:05am)

Title: simple conditional probability qns
Post by Valerfore on Oct 29^th, 2004, 6:05am

http://joshp.typepad.com/15/2003/11/conditional_pro.html

1. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. What is the chance that at least one child is a girl?

Replay with slight change

2. You are at the park, pushing your daughter on a swing. You meet a woman with her daughter and start a conversation. She says she has two children. What is the chance that both children are girls?

Rewind, again

3. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?

my answer is 3/4, 1/2, 1/2, but some people argued that the answer for qn 3 is 1/3.. experts pls help! :P

Title: Re: simple conditional probability qns
Post by TimK on Oct 29^th, 2004, 6:21am

They're right - the answer to question 3 is 1/3. There are three possibilities - GG, GB, BG, and in one of the 3, both children are girls.

Title: Re: simple conditional probability qns
Post by Grimbal on Oct 29^th, 2004, 8:14am

As I always say, probabilities can not be deduced from a situation, only from a process that brings to that situation. You have to know why the thing that did not happen didn't.

You might assume equal probabilites of having a girl or a boy. But if the woman showed up with a girl, you have to know why not a boy. If she picked a child randomly, there are situations where there was a girl in the family but she showed up with a boy. But these situations are not considered when computing the probability of 2 girls. So, showing up with a girl is not equivalent to answering yes to the question whether she has at least a girl.

If you ask about a girl, you might assume that of the 4 equiprobable cases, 1 of them is eliminated. In that case, all the situations where there is a girl in the family must be considered. The cases are BG, GB and GG, and the probability of 2 girls is 1/3.

Title: Re: simple conditional probability qns
Post by Emul P Edmon on Oct 30^th, 2004, 11:22am

Right, and the answer to scenario 2 is 1/4 not 1/2. Here are the truth tables

Scenario 1:
B B false
B G true
G B true
G G true
So the odds are 3/4

Scenario 2:
B B false
B G false
G B false
G G true
So the odds are 1/4

Scenario 3:
B G false
G B false
G G true
So the odds are 1/3

Title: Re: simple conditional probability qns
Post by Sir Col on Oct 30^th, 2004, 11:40am

Not quite, Emul P Edmon. In the second scenario she arrives with one of her children, which happens to be a girl. Without any further information, and as Grimbal pointed out, we must assume that the gender of the other child is independent. So P(GG)=1/2.

Here's another subtle variation...

4. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She tells you that she has two children and goes on to say, "I am so proud of my eldest, she recently won a poetry competition." What is the chance that both children are girls?

Title: Re: simple conditional probability qns
Post by Grimbal on Oct 30^th, 2004, 1:25pm

5. You are at the park, pushing your daughter on a swing. You meet a chinese woman and start a conversation. She tells you that she has two children and goes on to say, "I plan to send my eldest daughter to the best university I can get". What is the chance that both children are girls?

PS: just joking.

Title: Re: simple conditional probability qns
Post by rmsgrey on Oct 30^th, 2004, 1:49pm

It depends for both 4 and 5 on what you think the woman would have said in each of the possible cases.

For 4, if you think the woman picked a girl to talk about, then it's a 1/3 chance of both girls. If you think the choie of which child to mention was independent of gender, then it's a 1/2 chance of both girls.

For 5, you need to judge the probability that the woman would talk about her eldest daughter if she only had one daughter. Thinking about it, you can't assume that the probability she'd talk about her eldest daughter with 2 sons is 0 - she may not have a daughter yet, but may be planning to send the first daughter she subsequently has to a good college.

Title: Re: simple conditional probability qns
Post by mistysakura on Nov 6^th, 2004, 4:19pm

Lol to number 5. I'm guessing [hide]Pr (GG) = 1.
First reason: If there wasn't another daughter, she wouldn't have referred to her as the "eldest" daughter, but just "my daughter".

Second reason: If there was a son, she'd hardly be planning on sending her daughter to the best university, would she? (just kidding.)[/hide]

Title: Re: simple conditional probability qns
Post by THUDandBLUNDER on Nov 6^th, 2004, 8:25pm

Quote:

If there wasn't another daughter, she wouldn't have referred to her as the "eldest" daughter...

That's right, if she hadn't been Swiss-Chinese she would have referred to her as the elder daughter. :D

Title: Re: simple conditional probability qns
Post by Emul P Edmon on Nov 8^th, 2004, 4:21pm

Thank you, Sir Col, I had missed that. From now on I'll have to read more carefully. Thanks again...

Title: Re: simple conditional probability qns
Post by THUDandBLUNDER on Nov 10^th, 2004, 9:28pm

FRAT HOUSE? : You visit a house wherein a pair of twins live alone. You know that they are either boy/girl fraternal twins or identical girl twins. A girl answers the door. What is the probability that her sibling is also a girl?

Title: Re: simple conditional probability qns
Post by Sir Col on Nov 11^th, 2004, 12:42am

What a delightful variation...

::[hide]I shall assume that there is equal chance of the pair of twins being (1) B/G or (2) G1/G2.

There are four combinations of people answering the door, leaving the other person inside:
1B ... G
1G ... B
2G1 ... G2
2G2 ... G1

As a girl answered the door we are dealing with the last three possibilities, each with equal chance. So P(other twin is a girl)=2/3.
[/hide]::

Title: Re: simple conditional probability qns
Post by THUDandBLUNDER on Nov 11^th, 2004, 4:19am

Quote:

What a delightful variation...

But nevertheless varying much more, and somewhat less delightfully, than you have heretofore realized... :P

Quote:

I shall assume that there is equal chance of the pair of twins being (1) B/G or (2) G1/G2.

In this case is that a justified assumption?

Title: Re: simple conditional probability qns
Post by towr on Nov 11^th, 2004, 7:47am

on 11/11/04 at 04:19:50, THUDandBLUNDER wrote:

In this case is that a justified assumption?

Not in general, intuitively fraternal would be more common. But it's hard to estimate without some statistical data on the subject.
Even where they were born can enter into the vacation, seeing as in some places identical twins are unusually common.

Title: Re: simple conditional probability qns
Post by rmsgrey on Nov 11^th, 2004, 7:58am

For the general version of the frat house twins:
::[hide]
Assume the twins are identical with probability p, and that, if the twins are fraternal, the boy and the girl have equal chances of opening the door (which may not be justified):

so you have:
(G)B: (1-p)/2
(B)G: (1-p)/2
(G1)G2: p/2
(G2)G1: p/2

giving proability (p/2+p/2) / (p/2+p/2+(1-p)/2) = 2p/p+1 of the other sibling also being a girl
[/hide]::