I look at this as a way to start out 2009 and hopefully, the rest of my life: As a fresh start to making a better me.
A pic of me from 2007 finishing the race:
Link:
Wednesday, January 14, 2009
Signed up for the 2009 Shamrock Shuffle 8k run
After hearing that several friends will be doing the Shamrock Shuffle this year, I decided I wanted to challenge myself to it again. I did it two years ago, in 2007. I did little to no training for that, and ran at about a 13min/mile pace. I'm hoping to bring that down to a 10min/mile pace with some proper training.
I look at this as a way to start out 2009 and hopefully, the rest of my life: As a fresh start to making a better me.
A pic of me from 2007 finishing the race:
Link:
I look at this as a way to start out 2009 and hopefully, the rest of my life: As a fresh start to making a better me.
A pic of me from 2007 finishing the race:
Link:
Friday, January 02, 2009
Basic Combinatorial Mathematics
I absolutely love reading CodingHorror.com. It's easily one of my favorite blogs to read about problem solving from the viewpoint of a programming/developer. Recently, Jeff Atwood posted a problem:
Note: B = Boy, G = Girl
Some argued that the combinations to work off of is: BB, GG, BG (3 sets)
The full combination is: BB, GG, BG, GB (4 sets, where BG and GB are treated as separate sets)
The 3 set argument is that the BG, GB should be considered as one set, since order does not matter. But I beg to differ. Although order (older/younger) was not implied in the problem, it must be considered.
Here's why:
Having only 3 sets is basically saying that it's equally likely that a couple with two children will have either a Boy-Boy, Girl-Girl, or Boy-Girl set. This is not true. It is a 50/50 chance they will have children of the same gender (Boy-Boy or Girl-Girl versus a Boy/Girl combo). So a full 4 set combo must be used to deduce the answer to the problem.
The answer then, since we know that at least one is a girl, the BB set needs to get tossed out, leaving GG, BG, GB. So then 2 out of the 3 sets can have boys, so there is a 2/3 chance that the couple's second child is a boy.
Links:
Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?I was surprised by the number of readers who responded who got it wrong, very wrong. Quite a few who commented answered one-half, 50%. Reasoning ranged from it's always half, or because the ordering doesn't matter-- and that there are only 3 sets to choose from. The latter of which I want to discuss.
Note: B = Boy, G = Girl
Some argued that the combinations to work off of is: BB, GG, BG (3 sets)
The full combination is: BB, GG, BG, GB (4 sets, where BG and GB are treated as separate sets)
The 3 set argument is that the BG, GB should be considered as one set, since order does not matter. But I beg to differ. Although order (older/younger) was not implied in the problem, it must be considered.
Here's why:
Having only 3 sets is basically saying that it's equally likely that a couple with two children will have either a Boy-Boy, Girl-Girl, or Boy-Girl set. This is not true. It is a 50/50 chance they will have children of the same gender (Boy-Boy or Girl-Girl versus a Boy/Girl combo). So a full 4 set combo must be used to deduce the answer to the problem.
The answer then, since we know that at least one is a girl, the BB set needs to get tossed out, leaving GG, BG, GB. So then 2 out of the 3 sets can have boys, so there is a 2/3 chance that the couple's second child is a boy.
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