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

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b14kumar
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A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   20 Sep 2007, 15:35
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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
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C
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   20 Sep 2007, 22:59
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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

Add both of dem
1/8 + 1/8 = 1/4
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   20 Sep 2007, 23:29
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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


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

I suck at probability questions so I'm unsure about this answer.
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   21 Sep 2007, 02:31
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
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   21 Sep 2007, 03:17
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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
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Re: Tricky probability! [#permalink] New post   Updated on: 27 Aug 2008, 09:52
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.
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Re: Tricky probability! [#permalink] New post   27 Aug 2008, 09:43
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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
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Re: Tricky probability! [#permalink] New post   27 Aug 2008, 11:44
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2*C(4,3)*(1/2)^4 = 2*4*(1/2)^4 = 1/2 -> C
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Re: Tricky probability! [#permalink] New post   27 Aug 2008, 14:49
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
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Re: Tricky probability! [#permalink] New post   27 Aug 2008, 14:58
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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
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Re: Tricky probability! [#permalink] New post   30 Aug 2008, 16:09
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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
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Re: Tricky probability! [#permalink] New post   31 Aug 2008, 07:59
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.
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Re: A drawer holds 4 red hats and 4 blue hats. What is the proba [#permalink] New post   25 Oct 2013, 05:32
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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 :)
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   25 Aug 2018, 17:19
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
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   02 Feb 2020, 12:33
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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
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Re: A drawer holds 4 red hats and 4 blue hats. What is the probability of [#permalink] New post   02 Jan 2024, 11:54
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