Card picking dependant probability without replacement - P(6,6,Red)

#1
Hi All,

I'm trying to teach myself statistics online @ khan academy and although its a wonderful resource I have no one to turn to for help when I don't understand. I hope someone here might be able to give me some advice.

I've made some questions up myself to try and work out the answer and check it against a simulation i've written in VBA/Excel to really understand it all.

But.... I'm really stuck on working out the probability of picking card #1 with value 6, #2 with value 6 then #3 a Red card.

Working out P(6,6) is easy for me = 4/52 * 3/51 ≈ 0.4525%

My issue is that after having picked two 6's in a row, what is the probability of then picking out a red card - I just can't get my head around it. Because you could have already picked 2 red cards which would reduce the probability or you could have picked 1 or none.

I don't know how you work it out from here, and googling it doesn't help as i'm not wording my question right and not getting the results I want.

Can anyone help me / push me in the right direction.

ANYTHING would be great.

Thanks

Lewis
 
#2
How about splitting it up and finding the probabilities of red6 red6 red, red6 black6 red, black6 red6 red, and black6 black6 red