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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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17 Feb 2010, 13:01
jeeteshsingh wrote: Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: 2/8*3/7*2/6*3/5*2/4*5!/2!*2!=9/28=32.14% OR different way of counting: 2C1*3C2*3C2/ 8C5=9/28
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by 5!/2!2!, because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. Does the order matter in this question??? I dont think so.. as the question says... wat is the probability of getting 1 red, 2 green and 2 blue... and says nothing of the arrangement. Please comment. No, order does not matter here and nowhere I mention in my solution that it does. The point is that, in both cases, combination of RGGBB can occur in 5!/2!2! # of ways (RGGBB, GGRBB, BRBGG, ... total 30 combinations), so that's why we are multiplying by it.
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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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28 Feb 2010, 14:45
GMAT TIGER wrote: Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8% Why do you have 5!/2!2!? Does order matter? No... It matters in the sense that you have to sum all the different ways in which to choose the balls. BUNUEL's answer is correct. You have to multiply by 5!/(2!2!1!) because that is the number of ways to select 5 balls of which 2 are green (i.e. are similar), 2 are blue (i.e. are similar) and 1 is red. look at it this way... Try to find the probability of all the different ways in which you can draw 5 balls in the manner suggested. P(R,G,G,B,B) = 2/8*3/8*3/8*3/8*3/8 or P(G,R,G,B,B) = 2/8*3/8*3/8*3/8*3/8 or P(G,G,R,B,B) = 2/8*3/8*3/8*3/8*3/8 or . . . Thus the desired probability is = P(R,G,G,B,B) + P(G,R,G,B,B) + .... Since there are 5!/(2!2!) similar addends, the desired probability is 5!/(2!2!1!) * 2/8*3/8*3/8*3/8*3/8 = 5!/(2!2!) * (1/4) * (3/8)^2 * (3/8)^2



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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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10 Aug 2010, 17:45
Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)
OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. "The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8." Bunuel, About the above statement, can you explain the way of calculating Case I this way  using the counting approach. How would the calculation be different with the WITH case. Also as a general doubt when one says 5 balls are drawn randomly WITH replacement  drawing 5 balls randomly suggests I grab 5 directly, however with the WITH replacement clause, it seems to indicate that I draw one then put it back, then draw another and then replace it, essentially drawing balls 1 by 1.. Is this understanding correct?



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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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11 Aug 2010, 02:28
mainhoon wrote: Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)
OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. "The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8." Bunuel, About the above statement, can you explain the way of calculating Case I this way  using the counting approach. How would the calculation be different with the WITH case. Also as a general doubt when one says 5 balls are drawn randomly WITH replacement  drawing 5 balls randomly suggests I grab 5 directly, however with the WITH replacement clause, it seems to indicate that I draw one then put it back, then draw another and then replace it, essentially drawing balls 1 by 1.. Is this understanding correct? When we are told that 5 balls are drawn WITH replacement it means that we draw 1 ball and put it back and then another one and put it back and so on. GMAT wording would make it clear and unambiguous. So in the case WITH replacement we pick one ball at a time, so it's no need to use C here, as you wold have 1 from desired number divided 1 from all. For example instead of 2/8 you would have \(C^1_2/C^1_8\), which is complicated way of writing the same 2/8.
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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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18 Jan 2011, 09:54
Total ways of picking 5 out of 8 is 8C5=8!/5!3!=56 For what does follow next, we have to take into account that each time picked balls will be replaced: Probability of picking 2 out of 3 green balls is 3C2/56=3/56 Probability of picking 2 out of 3 blue balls is 3C2/56=3/56 Probability of picking 1 out of 2 red balls is 2C1/56=2/56
Inasmuch as picking green, blue and red balls are independent of each other because of replacement, we can add individual probabilities: (3C2/56)+( 3C2/56)+(2/56)=(3/56)+(3/56)+(2/56)=8/56=1/7≈14%
Here I would request participants of this forum to advice me whether my approach does work in all the cases with replacement when we have to do with probability?



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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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27 Sep 2013, 19:26
Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)
OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. Responding to a pm: @Himanshu Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR Here we are picking 5 balls RGGBB. How many different cases can we have? RGGBB RGBGB RBBGG BBGGR BGBGR and many more... The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.) NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before.
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Rich has 3 green, 2 red and 3 blue balls in a bag. He
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08 Dec 2015, 16:39
jeeteshsingh wrote: Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: 2/8*3/7*2/6*3/5*2/4*5!/2!*2!=9/28=32.14% OR different way of counting: 2C1*3C2*3C2/ 8C5=9/28
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by 5!/2!2!, because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. Does the order matter in this question??? I dont think so.. as the question says... wat is the probability of getting 1 red, 2 green and 2 blue... and says nothing of the arrangement. Please comment. For the counting method on Case II by Bunnuel, why are we not accounting for all the different combinations, just like in the probability example? What I mean is that for 2C1*3C2*3C2/ 8C5, unlike Bunnuel, I think that you need to multiply the numerator and the denominator by the number of combinations that are possible (since the order matters). For the probability method on Case II, we multiplied it by 5!/2!2!. Having said this, wouldn't we need to account for the number of combinations in the combination method? Thanks.



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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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22 Feb 2017, 02:31
VeritasPrepKarishma wrote: Responding to a pm: himanshuWhy are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR Here we are picking 5 balls RGGBB. How many different cases can we have? RGGBB RGBGB RBBGG BBGGR BGBGR and many more... The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.) NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before. Quote: I couldn figure out yet why dont we find simply the probability to select. For example selectin 2 green balls of 10 balls just 2/10. what is the difference here? dont it is asked simply to find prob of selecting these balls?
This is not correct. The probability of selecting 2 green balls out of 10 would not be 2/10. 2/10 would be the probability of selecting 1 green ball if out of 10 balls, exactly 2 were green. Probability = Favourable cases / Total cases Probability of selecting a green ball = 2/10 (if out of 10 balls, exactly 2 are green) Probability of selecting two green balls without replacement = 2/10 * 1/9 (for the first ball, the probability of it being green is 2/10 and the for the second ball, the probability of it being green is 1/9 because now we have only 9 balls left and 1 of them is green) Probability of selecting two green balls with replacement = 2/10 * 2/10 (each time the probability of picking a green ball is 2/10) Does this help?
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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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20 Dec 2018, 10:06
VeritasKarishma wrote: Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)
i am having a hard time wrapping my head around 5!2!/2! =14.8
OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. Responding to a pm: @Himanshu Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR Here we are picking 5 balls RGGBB. How many different cases can we have? RGGBB RGBGB RBBGG BBGGR BGBGR and many more... The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.) NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before. I am having a hard time figuring out why we do 2!*2! How does this get rid of the duplicates?



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Rich has 3 green, 2 red and 3 blue balls in a bag. He
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Updated on: 17 Dec 2019, 09:18
Hello Bunuel or any other expert, Could you guys please solve this problem by combination method for "with replacement" case? It would be very helpful to me. Thanks a lot.
Originally posted by a12bansal on 13 Dec 2019, 03:46.
Last edited by a12bansal on 17 Dec 2019, 09:18, edited 1 time in total.



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Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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17 Dec 2019, 09:02
Bunuel wrote: Let's clear out this one:
CASE I. Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)
CASE II. If the problem was WITHOUT replacement:
Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)
OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)
The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).
In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.
We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.
The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.
Hope now it's clear. Are we multiplying with 5! because the balls could have been picked in various color orders?




Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He
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