Overdetermined
Analytics, data, modeling and a healthy dose of cynicism.
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The great thing about ongoing series is that you can find all of them archived right here.
1. Building a Poll (Dirty D)
2. Building a Voter File (Blue Leader)
3. The Plebian's Guide to Polls (Student Redux)
4. Cool Tools (HummuSoft)
5. An Introduction to R (Pluribus)
Syndicate
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Edit: I tried to make the question easier to understand.
Here is the full question. Pretend that you have 4 pens in your pocket.
2 - Red
2 - Blue
Without looking, you remove a single pen. Then, without looking, you remove a second pen. What are the odds that the pens will be different colors? In other words, how likely is it that one pen will be blue and one pen will be red?
Think about it. It's both easier and harder to answer than you might think. The approach is IMPORTANT. If you think you know they answer, let me see "your work" in the comments section. This can be logic or you can just brute force it and solve it with a script. If you use a computer, I expect to see some code! I'm especially interested to see if anyone can come up with a way to solve this using SQL. I don't claim to be a SQL wizard, but it seemed like more trouble than it is worth but I could be wrong. I used R to check my answer.
This "poll" will be open for one week. When it closes, I will post my answer, and the code I have the supports my answer, and a miniature discussion on this topic. You get extra credit for using esoteric computer programming languages/tools.
If, by chance, you are the chemistry professor who introduced me to this little word problem at the gym, you are wrong. Badly Badly wrong. I should not have let your shiny PhD cloud my thinking. I was right and you were wrong.
Until next time.
Different Work, Same Answer
Agree on 2/3, but I come there through a simpler road.
Pick a pen. Any pen. Doesn't matter. Pr=1 for this event.
You have three pens left. Since you removed one, there's only one pen left that's the same color. Two of the remaining pens are of the other color. Therefore, Pr=2/3 that you pick the other color.
Incidentally, the CAPTCHA question I have to answer to post this comment is harder than the original probability question. (~_^)
Here's why 2/3
There are six conditions that can obtain: B1R1, B1R2, B1B2, B2R1, B2R2, and R1R2. Four out of these six satisfy the condition of having pens of different colors. Therefore, the answer is 2/3.
DD
Dirty D writes about polling, analytics, data and whatever else may cross his mind as being neat. Feel free to contact him by email : D I R T Y D AT O V E R D E T E R M I N E D DOT N E T.
How does this work, specifically?
Maybe I'm missing the point, but are you removing one at a time, or both at once?
DD
Dirty D writes about polling, analytics, data and whatever else may cross his mind as being neat. Feel free to contact him by email : D I R T Y D AT O V E R D E T E R M I N E D DOT N E T.
I should have been more
I should have been more clear. I am selecting them one at a time, but that may or may not be an important piece of information.