Kimmy’s Garden

OK, so Tuesday and Thurday mornings I get up, prepare my tea, and head off to Stats class for 8h30. At 11h30 I have the lab for this course, two times a week. First day of Lab, and may I add that the TA is very sweet, asks us to work in groups. $#!+. This is what I am thinking. I am sitting there, logged in to a computer in the second row, and boom, what sits beside me is a cutsie female, on her cell phone, texting her friends or whatever, whining to the hotty guy beside her in a I-don’t-know-how-to-do-this-can-you-show-me sort of sound.

Now, first of all, how can I not stand females pretending to be dumb in order to get attention from a male. So annoying. But the fact that we are supposed to be working in groups - with this sort of a partner - no thanks. I am taking this course for pleasure, not for frustration. So next lab I decide to show up late, so that I can choose with whom I will work. And then we worked on an independant MS Excel project. (Sigh)

OK, next lab, must show up a bit late again so that I can choose again. Oops, now the sweet TA is asking me if the lab times are not working out for me because I seem to come in late … now I feel like an @$$. And I am ashamed, but I am sitting beside s/o who looks like a good lab partner. Ah! Never sit in the middle of a row, beside an empty seat so that s/o else can sit beside you, changing the pair-up possibilities. Crap. So, here I think that I have picked a great spot, and who sits beside me, cell-phone girl. And anyhow, the original person who I picked ends up on E-Bay trying to buy concert tickets during the lab - now I have duds on both sides of me. Ai ya!

So, here I am doing all of my Stats homework, almost pulling out my hair on Conditional Probability, what the #e!! was I thinking originally? And I get to the question: If five applicants are pulled from a stack of 30 résumés, and only 5 are really decent candidats, what are your chances of pulling at least one of the 5 best?
Lets replace applicants with 5 other Good Lab Partners:

30 lab attendees, 4 per group, and lets assume that I am a good one ;-b…
that leaves 29 total left for 3 spots … 4 other Good partners left
= C to the 3 of 29 is 2,375,100 possible combinations of 3 other lab partners. The probability of getting at least 1 of the 4 other Good Partners (g) would be …

A = made of mini- events = {e1, 1 of g; e2, 2 of g; e3, 3 of g}.
P(A)=P(e1)+P(e2)+P(e3)
P(e1)=g1 or g2 or g3 or g4 =(1/29*28/28*27/27)*4=(756/21924)*4
P(e2)=(g1&g2)+(g1&g3)+(g1&g4)+(g2&g3)+(g2&g4)+(g3&g4)=
P(e2) … = (1/29*1/28*27/27)*6=(27/21924)*6
P(e3)=(g1&g2&g3)+(g1&g2&g4)+(g1&g3&g4)+(g2&g3&g4)=
P(e3) … = (1/29*1/28*1/27)*4=(1/21924)*4
P(A)=0.13793+0.00739+0.00018=0.1455=14.6%

Actually, that’s not bad … of course, the odds and actual events are

so

not the same …

I better think about this more … confirm my calculations