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# A bag contains 3 white balls, 3 black balls & 2 red balls.

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Director
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A bag contains 3 white balls, 3 black balls & 2 red balls. [#permalink]

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10 Aug 2010, 16:41
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Difficulty:

55% (hard)

Question Stats:

43% (01:04) correct 57% (01:14) wrong based on 37 sessions

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A bag contains 3 white balls, 3 black balls & 2 red balls. One by one three balls are drawn out without replacement. What is the probability that the third ball is red?

A. 1/2
B. 1/4
C. 3/8
D. 5/28
E. 2/5

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
[Reveal] Spoiler: OA

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10 Aug 2010, 16:55
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?
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10 Aug 2010, 17:08
My attempt:

I did not get the OA, but here is my understanding...

For the third ball to be a red one, here are the combination.

NR -- Non red ball.
R -- Red ball.

NR -- NR -- R (Non red ball drawn followed by a non-red ball followed by a red ball.)
R -- NR -- R
NR -- R -- R

Case 1: NR -- NR -- R

(1/7) * (1/6) * (1) -- (1/42)

(In the above calculation, I have included one red ball since we need only one red ball to be drawn as the third one.)

Case 2: R -- NR -- R

(1/2)*(1/6)*(1) = (1/12)

Case 3: NR -- R -- R

(1/6) * (1/2) * (1) -- (1/12)

Adding all the probabilities -- (1/42) + (1/12) + (1/12)

Gives me (4/21) --- which is not an option at all

Any thoughts on what went wrong in my approach.
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10 Aug 2010, 17:14
The possibilities are
WWR=2/56
WBR=3/56
WRR=1/56
BWR=2/56
BBR=3/56
BRR=1/56
RBR=1/56
RWR=1/56

Add all of them together you get 14/56=1/4.

Is there are better way to do this problem than working out all the combinations?

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10 Aug 2010, 17:31
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?
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10 Aug 2010, 17:47
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.

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10 Aug 2010, 18:41
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?
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10 Aug 2010, 18:49
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.

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10 Aug 2010, 18:50
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Sorry. For got to mention P(XXX)

P(RRX) = 2/8 x 1/7 x 6/6 = 1/28
P(RXX) = 2/8 x 6/7 x 5/6 = 5/28
P(XRX) = 6/8 x 2/7 x 5/6 = 5/28
P(XXX) = 6/8 x 5/7 x 4/6 = 5/14

Add them together you get 21/28. 1-(21/28) = 1/4.

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10 Aug 2010, 19:23
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!
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10 Aug 2010, 19:58
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.
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11 Aug 2010, 01:20
XXR=6/8*5/7*2/6
RXR=2/8*6/7*1/6
XRR=6/8*2/7*1/6
XXR+RXR+XRR=(6*5*2+2*6*1+6*2*1)/(8*7*6)=(5+1+1)/(4*7)=1/4 thus B

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17 Apr 2012, 03:15
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.

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17 Apr 2012, 03:50
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Expert's post
ENAFEX wrote:
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.

Similar questions to practice:
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each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-105990.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
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Re: Probability Q   [#permalink] 17 Apr 2012, 03:50
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