| Topic Name: |
Susquehanna International Group |
| Message Name: |
okay |
| Date Posted: |
03/26/2004 |
| In Reply To: |
That's where I disagree, I think there are initially six cases, you have child A and child B. Child A can be a girl or boy, older or younger and have a sister or brother. Once you see the girl cross the street, you know child A is a girl. There are now four equally likely possibilities: Child A has an older brother, older sister, younger brother or younger sister. These four equally likely situations yield a probability of 1/2 that the other child is a boy.
Tell me where I'm wrong.
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| Message: |
I've been mulling over this for a while.
When you are told that a female is at the mailbox, you should translate that to mean at least one child is a female. MF, FM, and FF all satisfy this condition. This is why the probability of the other child being a female is 1/3.
If, on the other hand, you're told that the older child is a female, then just FF and FM satisfy this so the probability is 1/2.
Here's another slightly different problem:
Lets say there are four families in a neighborhood, each has two kids. Family 1 has MM, Family 2 has MF, Family 3 has FM, and family 4 has FF. You see one of the neighborhood children and she is a female. What's the probability this girl has a sister?
The solution to this mirrors the thought process you are using in the interview question. The answer to this one is 50%.
Here's another one that's tough to decipher. You have the choice between two envolopes. One envolope contains X amount of money, the other has twice that amount. You choose an envolope then are given the choice to swich. Should you switch?
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