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A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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29 Aug 2010, 08:06
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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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40
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Re: What is the Probability of Red Ball? [#permalink]
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Favorable outcomes :
WWR > P(WWR)= 3/8 * 2/7 * 2/6 = 12/336 WRR > P(WRR)= 3/8 * 2/7 * 1/6 = 6/336 RWR > P(RWR)= 2/8 * 3/7 * 1/6 = 6/336 BBR > P(BBR)= 3/8 * 2/7 * 2/6 = 12/336 BRR > P(BRR)= 3/8 * 2/7 * 1/7 = 6/336 RBR > P(RBR)= 2/8 * 3/7 * 1/6 = 6/336 WBR > P(WBR)= 3/8 * 3/7 * 2/6 = 18/336 BWR > P(BWR)= 3/8 * 3/7 * 2/6 = 18/336
So, P(Total) = 84/336 = 1/4 = 0.25



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Re: What is the Probability of Red Ball? [#permalink]
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In problems such as these, the answer is just the probability based on the initial population... so 2/8 and it would be the same answer regardless of the nth try, 1st, 5th, 8th...
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Re: What is the Probability of Red Ball? [#permalink]
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Re: What is the Probability of Red Ball? [#permalink]
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29 Aug 2010, 14:34
Excellent Bunuel! I have seen a couple of such nice, handful and timesavers scattered across different posts . Have you considered posting a sticky where you collect them at one place. That would be very helpful , also the GMatClub may add it to the math section of the IPhone app. Thanks Posted from GMAT ToolKit



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Re: What is the Probability of Red Ball? [#permalink]
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29 Aug 2010, 16:35
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Bunuel wrote: krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40
Thanks The initial probability of drawing red ball is 2/8. Without knowing the other results, the probability of drawing red ball will not change for ANY successive drawing: second, third, fourth... Similar problems: psprobabilitygmatmademelosemymarbles36830.html?hilit=initial%20drawprobabilityofpickinglast97015.html?hilit=initial%20probabilityprobabilityquestionpleasesolve90272.html?hilit=initial%20probability#p686028Hope it's clear. !!!!! went the long way  still under 2min RedOtherRed +OtherRR+OOR \(\frac{1}{28}+\frac{1}{28}+\frac{5}{28}=\frac{7}{28}=\frac{1}{4}\)



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Re: What is the Probability of Red Ball? [#permalink]
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30 Aug 2010, 05:54
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P(drawing a red ball) = 1P(not drawing a red ball ) => 16C1/8C1 = 1/4



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Re: What is the Probability of Red Ball? [#permalink]
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30 Aug 2010, 05:58
psychomath wrote: P(drawing a red ball) = 1P(not drawing a red ball ) => 16C1/8C1 = 1/4 IMO when someone masters the ability to use combinations and probability at the same time in a gmat question, he/she has reached the level to move to a different topic.... @psychomath nice solution...



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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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24 Jun 2012, 01:58
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krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40 Responding to a pm: The point is that the probability of picking a red will not depend on which draw it is. In the first draw, probability of picking a red is 2/8 = 1/4 Probability of picking a red in the second draw: 2 cases: Case 1: First draw is non red. Probability = (6/8)*(2/7) This is equal to (2/8)(6/7). Think about it: Probability of picking non red first and then red will be the same as probability of picking a red first and then a non red. Case 2: First draw is red. Probability = (2/8)*(1/7) This is the probability of picking a red first and then a red again. Total probability of second draw being red = (2/8)*(6/7) + (2/8)*(1/7) = 2/8 This is just the probability of picking a red first and then any ball (non red or red). Probability of picking ANY ball will be 1. Hence, the probability of picking a red in the second draw will be the same as the probability of picking a red in the first draw. Similarly probability of picking a red in any draw will be the same.
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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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26 Jun 2012, 05:47
Hi Karishma, Thank You for the explanation. Just curious to know, do we have any scenario where this phenomenon fails. Thanks VeritasPrepKarishma wrote: The point is that the probability of picking a red will not depend on which draw it is.
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Re: What is the Probability of Red Ball? [#permalink]
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29 Jun 2013, 04:01
Bunuel wrote: krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40
Thanks The initial probability of drawing red ball is 2/8. Without knowing the other results, the probability of drawing red ball will not change for ANY successive drawing: second, third, fourth... Similar problems: psprobabilitygmatmademelosemymarbles36830.html?hilit=initial%20drawprobabilityofpickinglast97015.html?hilit=initial%20probabilityprobabilityquestionpleasesolve90272.html?hilit=initial%20probability#p686028Hope it's clear. Bunuel, Why is it not 7c2 * 1 / 8c3 = 3/8? selecting any 2 without 1 red is 7c2....selecting the last red is 1...so fav outcomes = 7c2 * 1 total outcomes = 8c3 prob = 7c2/8c3 Where am I going wrong? please explain..
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Re: What is the Probability of Red Ball? [#permalink]
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29 Jun 2013, 04:43
maaadhu wrote: Bunuel wrote: krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40
Thanks The initial probability of drawing red ball is 2/8. Without knowing the other results, the probability of drawing red ball will not change for ANY successive drawing: second, third, fourth... Similar problems: psprobabilitygmatmademelosemymarbles36830.html?hilit=initial%20drawprobabilityofpickinglast97015.html?hilit=initial%20probabilityprobabilityquestionpleasesolve90272.html?hilit=initial%20probability#p686028Hope it's clear. Bunuel, Why is it not 7c2 * 1 / 8c3 = 3/8? selecting any 2 without 1 red is 7c2....selecting the last red is 1...so fav outcomes = 7c2 * 1 total outcomes = 8c3 prob = 7c2/8c3 Where am I going wrong? please explain.. Don't quite understand what you are doing there. What does "selecting any 2 without 1 red is 7c2" mean? Anyway, the question basically asks: if we put all 8 balls in a row, what is the probability that the third ball is red? Similar questions might help: aboxcontains3yellowballsand5blackballsonebyone90272.htmlabagcontains3whiteballs3blackballs2redballs100023.htmleachoffourdifferentlockshasamatchingkeythekeys101553.htmlif40peoplegetthechancetopickacardfromacanister97015.htmlnewsetofmixedquestions150204100.html#p1208473amedicalresearchermustchooseoneof14patientsto127396.html
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Re: What is the Probability of Red Ball? [#permalink]
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01 Jul 2013, 08:19
Bunuel, first 2 balls can be selected from any of [3 black + 3 white + 1 red (7 balls)]  so selecting any 2 balls from 7 balls is 7c2. probability of selecting the remaining red ball is 1. so fav outcomes = 7c2*1. probability = 7c2/8c3 = 3/8/ I know my thinking is definitely wrong. Can you please point out the defect in my thinking?
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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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11 Jul 2013, 20:35
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VeritasPrepKarishma wrote: krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40 Responding to a pm: The point is that the probability of picking a red will not depend on which draw it is. In the first draw, probability of picking a red is 2/8 = 1/4 Probability of picking a red in the second draw: 2 cases: Case 1: First draw is non red. Probability = (6/8)*(2/7) This is equal to (2/8)(6/7). Think about it: Probability of picking non red first and then red will be the same as probability of picking a red first and then a non red. Case 2: First draw is red. Probability = (2/8)*(1/7) This is the probability of picking a red first and then a red again. Total probability of second draw being red = (2/8)*(6/7) + (2/8)*(1/7) = 2/8 This is just the probability of picking a red first and then any ball (non red or red). Probability of picking ANY ball will be 1. Hence, the probability of picking a red in the second draw will be the same as the probability of picking a red in the first draw. Similarly probability of picking a red in any draw will be the same. Responding to a pm: Quote: Please tell me what is wrong with my thinking.
first 2 balls can be selected from any of [3 black + 3 white + 1 red (7 balls)]  so selecting any 2 balls from 7 balls is 7c2.
probability of selecting the remaining red ball is 1. (You are mixing combinations with probability.)
so fav outcomes = 7c2*1.
probability = 7c2/8c3 = 3/8
I know my thinking is definitely wrong.
Can you please point out the defect in my thinking? There are a lot of problems here. If you want to use combinations here, you can do this: Assuming all balls are distinct, Number of ways of selecting 3 balls one after another without replacement = 7*6*1*2 (Keep one red ball away for the third pick. This can be done in 2 ways. Now of the 7 remaining balls, pick 1 for the first pick and then another for the second pick.) Total number of outcomes = 8*7*6 Probability = (7*6*2)/(8*7*6) = 1/4
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Re: What is the Probability of Red Ball? [#permalink]
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07 Aug 2013, 07:44
maaadhu wrote: Bunuel,
first 2 balls can be selected from any of [3 black + 3 white + 1 red (7 balls)]  so selecting any 2 balls from 7 balls is 7c2.
probability of selecting the remaining red ball is 1.
so fav outcomes = 7c2*1.
probability = 7c2/8c3 = 3/8/
I know my thinking is definitely wrong.
Can you please point out the defect in my thinking? The big mistake is here that you ignored the phrase of "drawing without replacement", then your probability didn't exclude 1 ball picked up previously.



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Re: What is the Probability of Red Ball? [#permalink]
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07 Aug 2013, 07:44
maaadhu wrote: Bunuel,
first 2 balls can be selected from any of [3 black + 3 white + 1 red (7 balls)]  so selecting any 2 balls from 7 balls is 7c2.
probability of selecting the remaining red ball is 1.
so fav outcomes = 7c2*1.
probability = 7c2/8c3 = 3/8/
I know my thinking is definitely wrong.
Can you please point out the defect in my thinking? The big mistake is here that you ignored the phrase of "drawing without replacement", then your probability didn't exclude 1 ball picked up previously.



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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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07 Aug 2013, 08:56
krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40 the probability of the 3rd ball not to be Red = 6/8 so, the required probability = 1 6/8 = 2/8 = 0.25
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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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07 Mar 2014, 04:54
Asifpirlo wrote: krishan wrote: 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. 0.25 B. 0.15 C. 0.35 D. 0.45 E. 0.40 the probability of the 3rd ball not to be Red = 6/8 so, the required probability = 1 6/8 = 2/8 = 0.25 Not quite sure because after 2 draws there won't be 8 balls anymore no? Also, what happens if we draw a red in the first two? Then won't the probability of drawing a red in the third draw be zero? Cheers J



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A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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02 Sep 2014, 16:02
mainhoon wrote: In problems such as these, the answer is just the probability based on the initial population... so 2/8 and it would be the same answer regardless of the nth try, 1st, 5th, 8th... By "In problems such as these", you meant "Replacement Problems", right? Just want to be sure so that I can apply it for similar problems. : )



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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]
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