Last visit was: 21 May 2024, 16:13 It is currently 21 May 2024, 16:13
Toolkit
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Request Expert Reply

# A drawer holds 4 red hats and 4 blue hats. What is the probability of

SORT BY:
Tags:
Show Tags
Hide Tags
Senior Manager
Joined: 17 Sep 2005
Posts: 300
Own Kudos [?]: 397 [25]
Given Kudos: 0
Manager
Joined: 09 Jul 2007
Posts: 84
Own Kudos [?]: 600 [1]
Given Kudos: 0
Senior Manager
Joined: 09 Aug 2006
Posts: 350
Own Kudos [?]: 976 [4]
Given Kudos: 0
Manager
Joined: 13 Mar 2007
Posts: 170
Own Kudos [?]: 158 [0]
Given Kudos: 0
Location: Russia, Moscow
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
b14kumar wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?

(a) 1/8
(b) 1/4
(c) 1/2
(d) 3/8
(e) 7/12

Please explain.

- Brajesh

C as well

Pr=Probability to draw red hat=4/8=1/2
!Pr=Probability not to draw red hat=(1-1/2)=1/2

We draw four times, but there are only 3 red hats we need, therefore we have following possible combinations (4C3=4):
1) Pr*Pr*Pr*!Pr=1/16
or
2) Pr*Pr*!Pr*Pr=1/16
or
3) Pr*!Pr*Pr*Pr=1/16
or
4) !Pr*Pr*Pr*Pr=1/16
Probability of drawing 3 red hats in four drawings=(1/16)*4 =1/4

Similarly probability of drawing 3 blue hats in four drawings=(1/16)*4 =1/4

Total probability=1/4+1/4=1/2
Senior Manager
Joined: 14 Jan 2007
Posts: 314
Own Kudos [?]: 906 [5]
Given Kudos: 0
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
1
Kudos
3
Bookmarks
There are 4 red and 4 blue hats.
Probability of drawing a red hat is 4/8 = 1/2 so for the blue hat.
Required probability for RRRB +BBBR,
There will be 4 ways to get exactly 3 reds - BRRR, RBRR, RRBR, RRRB
Same way there will be 4ways to get exactly blue.

So probability of getting exactly 3 reds and 3 blue = 4*1/2*1/2*1/2*1/2 + 4*1/2*1/2*1/2*1/2 = 1/2
Manager
Joined: 15 Jul 2008
Posts: 72
Own Kudos [?]: 396 [0]
Given Kudos: 0
Re: Tricky probability! [#permalink]
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

I think it should be 1/2 or 1/4. Slightly confused between the two.

With replacement, the prob of getting either color at each turn is 1/2. You want 3 out of 4 to be of the same color (say red). It becomes a binomial distribution problem.

probability of exactly 3 reds of out of 4 p = 4C3* (1/2)^3 * (1/2) = 1/4.

Now i am confused if i should multiply this by 2 to get the same possibility for blue as well.
I think i should but am not sure.

Originally posted by bhushangiri on 27 Aug 2008, 09:29.
Last edited by bhushangiri on 27 Aug 2008, 09:52, edited 1 time in total.
Director
Joined: 07 Nov 2007
Posts: 718
Own Kudos [?]: 3079 [1]
Given Kudos: 5
Location: New York
Re: Tricky probability! [#permalink]
1
Kudos
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

RRRB + BBBR
= 4!/3! * (1/2*1/2*1/2*1/2*) + 4!/3! * (1/2*1/2*1/2*1/2*)
= 1/4 +1/4 =1/2
Senior Manager
Joined: 01 Jan 2008
Posts: 258
Own Kudos [?]: 327 [1]
Given Kudos: 1
Re: Tricky probability! [#permalink]
1
Kudos
2*C(4,3)*(1/2)^4 = 2*4*(1/2)^4 = 1/2 -> C
VP
Joined: 29 Aug 2007
Posts: 1020
Own Kudos [?]: 1728 [0]
Given Kudos: 19
Re: Tricky probability! [#permalink]
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

first draw:

first hat could be anyhthing so prob = 1
second hat could be anyhthing so prob is again = 1
third hat could be the same as either the first one or the second one. so prob = 1/2
fourth hat has to be the same as the third one. so prob = 1/2

so it is: (1/1 x 1/1 x 1/2 x 1/2) = 1/4

second draw:

first hat could be anyhthing so prob = 1
second hat must be the color that was only one in first draw. so prob = 1/2
third hat has to be the same as the second one above. so prob = 1/2
fourth hat has to be the same as second/third ones. so prob = 1/2

so total prob = 1/4 + 1/8 = 3/8
Manager
Joined: 14 Jun 2008
Posts: 113
Own Kudos [?]: 151 [1]
Given Kudos: 0
Re: Tricky probability! [#permalink]
1
Kudos
due to symmtery, if we do it for Red hats, same probabiliyt of drawing out blue

to draw one Blue, and three Red i.e. BRRR
P = (1/2)^4

Similarily, for RBRR, RRBR, RRRB, probabilty = (1/2)^4

therefore probability of getting one blue and three reds
= (1/2)^4 * 4
= (1/2)^2

from symmetry, probability of drawing out three blue, and one red = (1/2)^2

therefore total = (1/2)^2 + (1/2)^2
= 1/2
Manager
Joined: 28 Jul 2004
Posts: 81
Own Kudos [?]: 243 [1]
Given Kudos: 2
Location: Melbourne
Concentration: General
Schools:Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth
Q50  V34 GMAT 2: 720  Q50  V38
Re: Tricky probability! [#permalink]
1
Kudos
x2suresh wrote:
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

RRRB + BBBR
= 4!/3! * (1/2*1/2*1/2*1/2*) + 4!/3! * (1/2*1/2*1/2*1/2*)
= 1/4 +1/4 =1/2

Hi Suresh, what is this 4!/3! factor ? Can you please explain it a bit more. What I understand is probablity of drawing any hat would be 1/2 (because they are replaced). So why it is not 2*(1/2*1/2*1/2*1/2)?

Thanks
Director
Joined: 07 Nov 2007
Posts: 718
Own Kudos [?]: 3079 [0]
Given Kudos: 5
Location: New York
Re: Tricky probability! [#permalink]
krishan wrote:
x2suresh wrote:
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

RRRB + BBBR
= 4!/3! * (1/2*1/2*1/2*1/2*) + 4!/3! * (1/2*1/2*1/2*1/2*)
= 1/4 +1/4 =1/2

Hi Suresh, what is this 4!/3! factor ? Can you please explain it a bit more. What I understand is probablity of drawing any hat would be 1/2 (because they are replaced). So why it is not 2*(1/2*1/2*1/2*1/2)?

Thanks

RRRB --> can be arranged in 4!/3! =4 ways..

If we don't do that we will be missing other arrangements BRRR RBRR RRBR

I hope it is clear now.
VP
Joined: 06 Sep 2013
Posts: 1343
Own Kudos [?]: 2401 [1]
Given Kudos: 355
Concentration: Finance
Re: A drawer holds 4 red hats and 4 blue hats. What is the proba [#permalink]
1
Bookmarks
Nihit wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?

A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 7/12

This is basically an easy problem because it is symmetric.

So for first case would be (1/2)^4 (4C3)
Then for second case it would be the same because probabilities are symmetrical

So in total we have 2(1/2^4)(4C3) = 1/2

Hope it helps
Cheers!
J
Senior Manager
Joined: 26 Jun 2017
Posts: 319
Own Kudos [?]: 327 [0]
Given Kudos: 334
Location: Russian Federation
Concentration: General Management, Strategy
WE:Information Technology (Other)
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
b14kumar wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?

A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 7/12

Red hats equal to blue hats.

3 red hats.
The probability is 4/8*4/8*4/8*4/8 for 1 scenario of 3 red hats. (4 scenarios at all: 4C3 - 123, 124, 134, 234)
So the whole probabilty is 4*4/8*4/8*4/8*4/8

The same for blue hats.
So, the answer is 2*4*4/8*4/8*4/8*4/8 = 1/2 - option C
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18886
Own Kudos [?]: 22289 [0]
Given Kudos: 285
Location: United States (CA)
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
Expert Reply
b14kumar wrote:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?

A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 7/12

First, let’s determine the probability of selecting exactly 3 red hats (and 1 blue hat).
P(R-R-R-B) = (1/2)^4 = 1/16
Since R-R-R-B can be arranged in 4!/3! = 4 ways, the overall probability is 4/16 = 1/4.
Since we have the same number of red and blue hats, the probability of selecting exactly 3 blue hats (and 1 red hat) is also 1/4.
Thus, the probability of selecting exactly 3 red hats or 3 blue hats is 1/4 + 1/4 = 2/4 = 1/2.

Answer: C
Non-Human User
Joined: 09 Sep 2013
Posts: 33140
Own Kudos [?]: 829 [0]
Given Kudos: 0
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
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 drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink]
Moderator:
Math Expert
93373 posts