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Ok so my brain is not quite functioning right, but here is my thought.. If I found the probablity of pulling out both the red balls in the first 2 tries and subtracted the probablility of this from 1, that should leave me with the probability of having a red ball in the 3rd attempt (as I think all remaining combinations will have a R at the end)
P(R in 3rd attempt) = 1 - P(RR W/B)
Here is where I am confused. It says without replacement, ok so
P(RR W/B) = 2/8 (2Red in 8) x 1/7 (1Red left in 7) x 6/6 (Dont care it will be W or B) = 1/28
P(R in 3rd attempt) = 1-1/28 = 27/28, Of course there is no answer choice, just my luck. Can someone enlighten please.. Bunuel?
I see you are listing the combinations. If you listed all the combinations where in RRX (X=W/B) how many do we get? lets see RRW = (2/8 x 1/7 x 3/6) = 1/56 RRB = 1/56 So (XYR) = 1-1/56-1/56 = 27/28 Goes back to my posting above.. Where is the mistake?
You are saying (1 - possibility of RRX). But I think you also have to take into consideration the following combinations: RXX or XRX. Because these combinations won't give you a Red on the third pick, they should also be subtracted from 1.
Good point. So I should be looking at the combinations RRX RXX XRX that should cover all So for P(RRX) = 2/8 x 1/7 x 6/6 = 2/56 P(RXX) = 3/8 x 6/7 x 5/6 = 15/56 P(XRX) = 6/8 x 2/7 x 5/6 = 10/56 I only got upto 2+15+10=27 so that is 1-27/56, I am missing 15 more cases so that 1-42/56 = 14/56=1/4 where did I go wrong now?
In terms of probability, the distribution at any time during the extraction process will reflect the initial distribution (this is because we are considering probability), so the answer is given by the initial distribution: 2/8=1/4.
Excellent swdatta! P(XXX) how could I miss that! Going by what toshio is saying though, it seems the white and black are all misleading, all that matters is the initial probability.. so if that is really true, then we just calculate the initial value and are done!
the approach by toshio86 is better. I was going to post this approach but re-read the full thread and found it to be mentioned already. My thought process was: since we dont dont know what was drawn out, the probability remains the same (even with or without replacement), unless any other skew was mentioned in the question. Why this approach is better is because it does not depend on sequence or the number of ball drawn out (1st, 3rd or 8th etcetera). Imagine doing the selections for the fifth ball drawn and getting the same answer. Time saved = Better score. Cheers
I tried it using combinations and I got the answer wrong.
I used (6C2/8C2)*(2C1/6C1).
(6C2/8C2) - selection of first two balls (2C1/6C1) - selection of one red ball from the remaining.
The simplest solution of this question is as follows: since the initial probability of drawing red ball is 2/8, then (without knowing the other results) the probability of drawing red ball will not change for ANY successive drawing: second, third, fourth... and will also equal to 2/8.
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