GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Aug 2020, 04:41 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # A drawer holds 4 red hats and 4 blue hats. What is the probability of

Author Message
TAGS:

### Hide Tags

Senior Manager  Joined: 17 Sep 2005
Posts: 363
A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

1
16 00:00

Difficulty:   95% (hard)

Question Stats: 45% (02:07) correct 55% (02:03) wrong based on 231 sessions

### HideShow timer Statistics

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
Manager  Joined: 09 Jul 2007
Posts: 108
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

B..

Probability of selectin 3 red hats
4c1/8c1 x 4c1/8c1 x 4c1/8c1 = 1/8

Probability of selectin 3 blue hats
4c1/8c1 x 4c1/8c1 x 4c1/8c1 = 1/8

1/8 + 1/8 = 1/4
Director  Joined: 09 Aug 2006
Posts: 526
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

3
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

- Brajesh

I get C.

Prob. of A happening exactly k times in n repeated trials is:

C(n,k) * P^k * (1-p)^n-k

For red hats:
n = 4
k = 3
p = 4/8 = 1/2
1-p = 1-1/2 = 1/2

P(exactly 3 red hats) = 4!/3! * (1/2)^3 * (1/2)^1 = 1/4

Same way, p(exactly 3 blue hats) = 1/4

total p = 1/4+1/4 = 1/2

Manager  Joined: 13 Mar 2007
Posts: 213
Location: Russia, Moscow
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

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

- 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: 497
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

1
1
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: 147

### Show Tags

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, 08:29.
Last edited by bhushangiri on 27 Aug 2008, 08:52, edited 1 time in total.
VP  Joined: 07 Nov 2007
Posts: 1060
Location: New York

### Show Tags

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: 418

### Show Tags

1
2*C(4,3)*(1/2)^4 = 2*4*(1/2)^4 = 1/2 -> C
SVP  Joined: 29 Aug 2007
Posts: 1712

### Show Tags

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: 137

### Show Tags

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: 107
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth

### Show Tags

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
VP  Joined: 07 Nov 2007
Posts: 1060
Location: New York

### Show Tags

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: 1491
Concentration: Finance
Re: A drawer holds 4 red hats and 4 blue hats. What is the proba  [#permalink]

### Show Tags

1
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  P
Joined: 26 Jun 2017
Posts: 380
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]

### Show Tags

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 V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11361
Location: United States (CA)
Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of  [#permalink]

### Show Tags

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.

_________________

# Scott Woodbury-Stewart | Founder and CEO | Scott@TargetTestPrep.com

250 REVIEWS

5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE

NOW WITH GMAT VERBAL (BETA)

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of   [#permalink] 02 Feb 2020, 11:33

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