Welcome! Check out our free B-School Guides to learn how you compare with other applicants.

Probablity ques

This topic has 5 expert replies and 9 member replies
selango GMAT Titan
Joined
29 Dec 2009
Posted:
1460 messages
Followed by:
3 members
Thanked:
133 times
Probablity ques Thu Jun 24, 2010 11:19 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A.1/4
B.1/2
C.5/8
D.2/3
E.3/4

OA C

Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
singhpreet1 Rising GMAT Star
Joined
08 Jun 2010
Posted:
95 messages
Thanked:
7 times
Thu Jun 24, 2010 11:52 pm
selango wrote:
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A.1/4
B.1/2
C.5/8
D.2/3
E.3/4

OA C
took me a while to get my head around this one.

the fixed probability is only on the fourth ball which should be black, no restriction on the rest: therefore: if the fourth must be black we start with only 7 balls to choose from and it is without replacement so 1 ball is reduced with every ball we take out.:

1st ball= 7/8, second ball= 6/7, third ball=5/6, fourth ball 5/5, fourth ball has a chance of being any of the 5 black balls out of 5 the remaining 5 balls.

7/8*6/7*5/6*5/5 which gives u 5/8, option C

kvcpk GMAT Titan
Joined
31 May 2010
Posted:
1874 messages
Followed by:
4 members
Thanked:
197 times
Fri Jun 25, 2010 12:27 am
Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.

GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
6093 messages
Followed by:
1039 members
Thanked:
5259 times
GMAT Score:
790
Fri Jun 25, 2010 1:20 pm
kvcpk wrote:
Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.
Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!

_________________
Mitch Hunt
GMAT Private Tutor and Instructor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me about long distance tutoring!

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

GMAT/MBA Expert

kevincanspain GMAT Instructor
Joined
22 Mar 2007
Posted:
578 messages
Followed by:
46 members
Thanked:
132 times
GMAT Score:
790
Fri Jun 25, 2010 3:19 pm
This is so because each ball is equally likely to be the fourth one (or nth one) drawn

_________________
Kevin Armstrong
GMAT Instructor
Gmatclasses

Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today!
kvcpk GMAT Titan
Joined
31 May 2010
Posted:
1874 messages
Followed by:
4 members
Thanked:
197 times
Fri Jun 25, 2010 10:12 pm
GMATGuruNY wrote:
kvcpk wrote:
Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.
Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!
Thanks for sharing it Mitch... Very Useful

missrochelle Really wants to Beat The GMAT!
Joined
09 Jun 2010
Posted:
117 messages
Wed Aug 04, 2010 1:31 pm
GMATGuruNY wrote:
kvcpk wrote:
Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.
Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!
Using this logic, why would the other problem be solved this way:
P(black first pick) 5/8
P (black 2nd pick) 4/7
P (black 3rd pick) 3/6
P (black 4th pick) 2/5
I think you'd have to add the first three and then multiply the black 4th pick ... but this gets really complicated. I guess I'm just trying to figure out how to do the two problems the same way, since we didnt use this methology to anwser the yellow /black ball problem.

Gurpinder GMAT Destroyer!
Joined
14 Dec 2009
Posted:
659 messages
Followed by:
2 members
Thanked:
27 times
Test Date:
Target GMAT Score:
GMAT Score:
Wed Aug 04, 2010 2:36 pm
singhpreet1 wrote:
took me a while to get my head around this one.

the fixed probability is only on the fourth ball which should be black, no restriction on the rest: therefore: if the fourth must be black we start with only 7 balls to choose from and it is without replacement so 1 ball is reduced with every ball we take out.:

1st ball= 7/8, second ball= 6/7, third ball=5/6, fourth ball 5/5, fourth ball has a chance of being any of the 5 black balls out of 5 the remaining 5 balls.

7/8*6/7*5/6*5/5 which gives u 5/8, option C
why is the 1st pick 7/8. shouldnt it be 8/8 because anything is a favorable outcome since there is no restriction.

missrochelle Really wants to Beat The GMAT!
Joined
09 Jun 2010
Posted:
117 messages
Wed Aug 04, 2010 5:07 pm
it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.

GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
6093 messages
Followed by:
1039 members
Thanked:
5259 times
GMAT Score:
790
Thu Aug 05, 2010 3:00 am
missrochelle wrote:
it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.
Here are all the independent probabilities that would get us a good outcome (black on the 4th pick):

P(BBBB) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

P(YBBB) = 3/8 * 5/7 * 4/6 * 3/5 = 3/28
Since (BYBB) and (BBYB) are also good outcomes, total number of ways to get 1 yellow with black on the 4th pick = 3 * 3/28 = 9/28

P(YYBB) = 3/8 * 2/7 * 5/6 * 4/5 = 1/14
Since (BYYB) and (YBYB) are also good outcomes, total number of ways to get 2 yellow with black on the 4th pick = 3 * 1/14 = 3/14

P(YYYB) = 3/8 * 2/7 * 1/6 * 5/5 = 1/56

Since all of the above are good outcomes, we add the fractions:

1/14 + 9/28 + 3/14 + 1/56 = 35/56 = 5/8.

Much easier just to remember the rule I posted above:

P(B) on the nth pick is the same as P(B) on the 1st pick: 5/8.

_________________
Mitch Hunt
GMAT Private Tutor and Instructor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me about long distance tutoring!

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
viju9162 GMAT Destroyer!
Joined
11 Jun 2007
Posted:
434 messages
Thanked:
5 times
Target GMAT Score:
750+
GMAT Score:
600
Thu Aug 05, 2010 9:41 am
GMATGuruNY wrote:
missrochelle wrote:
it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.
Here are all the independent probabilities that would get us a good outcome (black on the 4th pick):

P(BBBB) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

P(YBBB) = 3/8 * 5/7 * 4/6 * 3/5 = 3/28
Since (BYBB) and (BBYB) are also good outcomes, total number of ways to get 1 yellow with black on the 4th pick = 3 * 3/28 = 9/28

P(YYBB) = 3/8 * 2/7 * 5/6 * 4/5 = 1/14
Since (BYYB) and (YBYB) are also good outcomes, total number of ways to get 2 yellow with black on the 4th pick = 3 * 1/14 = 3/14

P(YYYB) = 3/8 * 2/7 * 1/6 * 5/5 = 1/56

Since all of the above are good outcomes, we add the fractions:

1/14 + 9/28 + 3/14 + 1/56 = 35/56 = 5/8.

Much easier just to remember the rule I posted above:

P(B) on the nth pick is the same as P(B) on the 1st pick: 5/8.
Very well explained!!!

_________________
"Native of" is used for a individual while "Native to" is used for a large group

nox104 Rising GMAT Star
Joined
09 Sep 2008
Posted:
69 messages
Thanked:
3 times
Target GMAT Score:
700+
Thu Aug 05, 2010 5:52 pm
Quote:
P(B) on the nth pick is the same as P(B) on the 1st pick
Can you explain what this rule actually means in words and when is it applicable?

I followed an easier approach and arrived at the same answer.

First, fixing the 4th pick as a black, we have 7 balls (3 Y + 4 B) remaining. Using the ANAGRAMS technique,

P (B on the 4th) = all arrangements of YYYBBBB / all arrangements of YYYBBBBB

Numerator = 7!/3!4! = 35
Denominator = 8!/3!5! = 56

GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
6093 messages
Followed by:
1039 members
Thanked:
5259 times
GMAT Score:
790
Thu Aug 05, 2010 6:01 pm
nox104 wrote:
Quote:
P(B) on the nth pick is the same as P(B) on the 1st pick
Can you explain what this rule actually means in words and when is it applicable?

I followed an easier approach and arrived at the same answer.

First, fixing the 4th pick as a black, we have 7 balls (3 Y + 4 B) remaining. Using the ANAGRAMS technique,

P (B on the 4th) = all arrangements of YYYBBBB / all arrangements of YYYBBBBB

Numerator = 7!/3!4! = 35
Denominator = 8!/3!5! = 56

Here's the rule:

P(X) on the nth pick = P(X) on the 1st pick.

So if we have 3 yellow and 5 black marbles:

P(B) on the 1st pick is 5/8
P(B) on the 2nd pick is 5/8
P(B) on the 3rd pick is 5/8
P(B) on the 4th pick is 5/8

P(B) is the same for each pick.

Does this help?

_________________
Mitch Hunt
GMAT Private Tutor and Instructor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me about long distance tutoring!

Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
nox104 Rising GMAT Star
Joined
09 Sep 2008
Posted:
69 messages
Thanked:
3 times
Target GMAT Score:
700+
Thu Aug 05, 2010 6:06 pm
Well I figured that it would be 5/8 each time from "P(B) on the nth pick is the same as P(B) on the 1st pick". But I was kinda trying to visualize it and understand why it is so..

GMAT/MBA Expert

GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
6093 messages
Followed by:
1039 members
Thanked:
5259 times
GMAT Score:
790
Wed Aug 11, 2010 6:35 am
nox104 wrote:
Well I figured that it would be 5/8 each time from "P(B) on the nth pick is the same as P(B) on the 1st pick". But I was kinda trying to visualize it and understand why it is so..

Algebraically:

P(B) on the 1st pick = B/(B+Y)

P(B) on the 2nd pick:
P(BB) = B/(B+Y) * (B-1)/(B+Y-1) = (B^2-B)/[(B+Y)(B+Y-1)]
P(YB) = Y/(B+Y) * B/(B+Y-1) = BY[(B+Y)(B+Y-1)]
Adding, we get P(B) on 2nd pick:
= (B^2-B)/[(B+Y)(B+Y-1)] + BY[(B+Y)(B+Y-1)]
= (B^2-B+BY)/[(B+Y)(B+Y-1)]
=[B(B-1+Y)]/[(B+Y)(B+Y-1)]
=B/(B+Y)
=P(B) on the 1st pick

Hence, P(B) on the 2nd pick = P(B) on the 1st pick.

From this result we can predict that P(B) on the nth pick = B/(B+Y) = P(B) on the 1st pick.

Does this help?

_________________
Mitch Hunt
GMAT Private Tutor and Instructor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me about long distance tutoring!

Thanked by: merlions777
Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

Best Conversation Starters

1 varun289 40 topics
2 guerrero 21 topics
3 sana.noor 20 topics
4 killerdrummer 19 topics
5 sanaa.rizwan 14 topics
See More Top Beat The GMAT Members...

Most Active Experts

1 Brent@GMATPrepNow

GMAT Prep Now Teacher

203 posts
2 GMATGuruNY

The Princeton Review Teacher

140 posts
3 Jim@StratusPrep

Stratus Prep

100 posts
4 Anju@Gurome

Gurome

99 posts