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**KrazyHorse** · February 23, 2011, 11:36

Write down conditional prob eqn for p(a drawn in a pair GIVEN b drawn in that pair). Now apply fact that draws are independent

**KrazyHorse** · February 23, 2011, 11:41

P(a|b) = P(a && b) / P(b)
P(b) = 1-(11/12)*(10/11)
P(a && b) = 2 * (1/12) *(1/11)

**MikeH** · February 23, 2011, 11:49

Thank you!

**MikeH** · February 23, 2011, 11:53

It's obvious when you see it.

**germanos** · February 23, 2011, 12:03

Liar

**MikeH** · February 23, 2011, 12:06

I didn't actually go to any of our maths lectures, but I did read the book before the exam.

**Wezil** · February 23, 2011, 12:07

I was going to post the solution but I figured I'd give KH a chance to see what he could come up with.

**Agent Sox** · February 23, 2011, 19:32

Oh wow so there are actually other math people on this forum? That's cool.

**Aeson** · February 23, 2011, 23:45

probability that mike has a man crush on one of the others and is hoping 4 some1 2 tell him the odds of getting matched up arent as bad as they seem?

**Perfection** · February 23, 2011, 23:50

If it was me, the probability would be 1 because I would rig the drawing.

**KrazyHorse** · February 23, 2011, 23:55

Originally posted by Agent Sox View Post

Oh wow so there are actually other math people on this forum? That's cool.

24....

**MikeH** · February 24, 2011, 04:28

I thought it was 500?

**KrazyHorse** · February 24, 2011, 08:33

Own goal

**MikeH** · February 24, 2011, 09:38

I must have missed something then.

Probability question.