Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 May 2016, 04:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A box contains 3 yellow balls and 5 black balls. One by one

Author Message
TAGS:

### Hide Tags

Manager
Joined: 27 Aug 2009
Posts: 144
Followers: 2

Kudos [?]: 19 [2] , given: 1

A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

### Show Tags

09 Feb 2010, 22:00
2
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:00) correct 22% (01:48) wrong based on 295 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Jun 2012, 02:28, edited 1 time in total.
Manager
Joined: 26 May 2005
Posts: 209
Followers: 2

Kudos [?]: 105 [1] , given: 1

### Show Tags

09 Feb 2010, 23:54
1
KUDOS
3 yellow 5 black
possible outcomes for the first 3 tries and 4th is black
yyyb - 3/8 * 2/7 * 1/6 * 5/5 = 30/(8*7*6*5)
yybb - 3/8 * 2/7 * 5/6 * 4/5 = 120/(8*7*6*5)
ybyb - 3/8 * 5/7 * 2/6 * 4/5 = 120/(8*7*6*5)
ybbb -3/8 * 5/7 * 4/6 * 3/5 = 180/(8*7*6*5)
bbbb - 5/8 * 4/7 * 3/6 * 2/5 = 120/(8*7*6*5)
bbyb - 5/8 * 4/7 * 3/6 * 3/5 = 180/(8*7*6*5)
bybb - 5/8 * 3/7 * 4/6 * 3/5 = 180/(8*7*6*5)
byyb - 5/8 * 3/7 * 2/6 * 4/5 = 120/(8*7*6*5)

total = 1050/(8*7*6*5)= 5/8
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5775

Kudos [?]: 70851 [8] , given: 9857

### Show Tags

10 Feb 2010, 05:56
8
KUDOS
Expert's post
3
This post was
BOOKMARKED
dmetla 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?

The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.
_________________
Senior Manager
Joined: 21 Jul 2009
Posts: 366
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Followers: 17

Kudos [?]: 140 [0], given: 22

### Show Tags

10 Feb 2010, 12:35
Bunuel wrote:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5775

Kudos [?]: 70851 [2] , given: 9857

### Show Tags

10 Feb 2010, 12:57
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
BarneyStinson wrote:
Bunuel wrote:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?
_________________
Manager
Joined: 27 Aug 2009
Posts: 144
Followers: 2

Kudos [?]: 19 [0], given: 1

### Show Tags

10 Feb 2010, 21:36
Thanks for the explanation. Answer is 5/8
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2797
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 217

Kudos [?]: 1416 [0], given: 235

### Show Tags

11 Feb 2010, 00:35
Bunuel wrote:
dmetla 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?

The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Nice explanation....Thanks
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Joined: 10 Feb 2010
Posts: 193
Followers: 2

Kudos [?]: 81 [0], given: 6

### Show Tags

12 Feb 2010, 07:49
I didnt realize it was so straight forward --
Manager
Joined: 29 Nov 2009
Posts: 107
Location: United States
Followers: 1

Kudos [?]: 30 [6] , given: 5

### Show Tags

16 Feb 2010, 10:38
6
KUDOS
I thought of it this way:

All possibilities with fourth ball fixed as black XXXBXXXX

7 spots, three filled by yellow, four filled by black:
(7!)/(3!*4!) = 35

All total possibilities of arranging 3 yellow and 5 black balls:
(8!)/(3!*5!) = 56

35/56 = 5/8
Senior Manager
Joined: 01 Feb 2010
Posts: 267
Followers: 1

Kudos [?]: 47 [0], given: 2

### Show Tags

16 Feb 2010, 11:19
Bunuel wrote:
BarneyStinson wrote:
Bunuel wrote:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Nice explanation didn't thought after reading questin that it was so simple.
Intern
Status: Looking for High GMAT Score
Joined: 19 May 2012
Posts: 37
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Internet and New Media)
Followers: 0

Kudos [?]: 6 [0], given: 58

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

### Show Tags

22 Jun 2012, 02:23
I got answer is 5/8 but choices are not displaying!
_________________

“The best time to plant a tree was 20 years ago. The second best time is now.” – Chinese Proverb

Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5775

Kudos [?]: 70851 [0], given: 9857

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

### Show Tags

22 Jun 2012, 02:28
Expert's post
MBACHANGE wrote:
I got answer is 5/8 but choices are not displaying!

_________________
Intern
Joined: 05 Sep 2011
Posts: 7
Followers: 0

Kudos [?]: 1 [0], given: 1

### Show Tags

26 Jun 2012, 05:39
Bunuel wrote:
BarneyStinson wrote:
Bunuel wrote:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

@Bunuel
please correct me if I am wrong

"if it is said that first card is sprade, then what is the probability of the 3rd card would be a sprade?-- in this case we wouldn't consider initial probability?"
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5775

Kudos [?]: 70851 [0], given: 9857

### Show Tags

27 Jun 2012, 02:06
Expert's post
marzan2011 wrote:
Bunuel wrote:
BarneyStinson wrote:
Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

@Bunuel
please correct me if I am wrong

"if it is said that first card is sprade, then what is the probability of the 3rd card would be a sprade?-- in this case we wouldn't consider initial probability?"

If the first card is spade then we are left with 4 spades and 3 hearts and the probability that 3rd (or any other card) will be a spade is 4/7.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9696
Followers: 466

Kudos [?]: 120 [0], given: 0

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

### Show Tags

12 Feb 2014, 22:21
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 38 [0], given: 23

### Show Tags

20 Apr 2014, 13:13
Bunuel wrote:
BarneyStinson wrote:
Bunuel wrote:
The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Hope it's clear.

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Hi Bunuel,

How can we say that the probability stays the same -- aren't there unknowns involved?

What I mean is, if the first ball was randomly selected to be Black, wouldn't that change the probability of the second ball being black to 4/7? What I mean by that is -- how is probability remaining constant if *some* card is being chosen, even if we don't know what the card being chose is?

I tried to follow this approach and was completely off -- can you please explain why? $$P=C\frac{8}{5}*\frac{5}{8} * \frac{3}{8}^7$$
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5775

Kudos [?]: 70851 [0], given: 9857

### Show Tags

21 Apr 2014, 01:02
Expert's post
russ9 wrote:
Bunuel wrote:
BarneyStinson wrote:

Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.

Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8.

Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change?

Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade?

I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?

Hi Bunuel,

How can we say that the probability stays the same -- aren't there unknowns involved?

What I mean is, if the first ball was randomly selected to be Black, wouldn't that change the probability of the second ball being black to 4/7? What I mean by that is -- how is probability remaining constant if *some* card is being chosen, even if we don't know what the card being chose is?

I tried to follow this approach and was completely off -- can you please explain why? $$P=C\frac{8}{5}*\frac{5}{8} * \frac{3}{8}^7$$

If we knew that the firs ball was black, then yes, the probability that the second ball will be black would be 3/7/ Nut notice that for our original question we don't know the results of 1st, 2nd and 3rd draws...

Hope that similar questions will help to understand better:
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html#p747637
new-set-of-mixed-questions-150204-100.html#p1208473
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9696
Followers: 466

Kudos [?]: 120 [0], given: 0

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

### Show Tags

04 May 2015, 21:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: A box contains 3 yellow balls and 5 black balls. One by one   [#permalink] 04 May 2015, 21:26
Similar topics Replies Last post
Similar
Topics:
A box contains 5 red & 6 yellow balls. If 4 balls are chosen at random 5 13 Dec 2014, 01:47
1 There are 3 black balls and 10 white balls. If one is to pick 5 balls, 4 21 Nov 2011, 11:27
A box contains 3 yellow balls and 5 black balls. One by one 2 08 Dec 2010, 19:20
2 A bag contains 3 white balls, 3 black balls & 2 red balls. 13 10 Aug 2010, 17:41
34 A box contains 3 yellow balls and 5 black balls. One by one 24 29 Sep 2009, 01:14
Display posts from previous: Sort by