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

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A drawer holds 4 red and 4 blue hats. What is the [#permalink]

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New post 16 Jul 2008, 20:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A drawer holds 4 red and 4 blue hats. What is the probability of getting exactly three read hats or three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

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Re: Probability [#permalink]

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New post 16 Jul 2008, 20:50
quangviet512 wrote:
A drawer holds 4 red and 4 blue hats. What is the probability of getting exactly three read hats or three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?


Probablity of taking out a particular color hat = 4/8 = 1/2

P(BBBR) = (1/2)^4
P(BBRB) = (1/2)^4
P(BRBB) = (1/2)^4
P(RBBB) = (1/2)^4
total = 4 * (1/2)^4

similary for (RRRB) = 4 * (1/2)4

answer = 8 * (1/2) ^4 = 1/2

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Re: Probability [#permalink]

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New post 17 Jul 2008, 12:53
2 choices for the first/second/third/fourth hat ==> Total number of possibilities : 2*2*2*2 = 16

4 ways to have 3 blue hats (BBBR, BBRB, BRBB, RBBB)

4 ways to have 3 red hats (RRRB, RRBR, RBRR, BRRR)

==> Answer is (4+4)/16 = 1/2

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Re: Probability [#permalink]

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New post 17 Jul 2008, 15:44
what about assuming, without replacement?
16/35?

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Re: Probability [#permalink]

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New post 18 Jul 2008, 11:42
young_gun wrote:
what about assuming, without replacement?
16/35?


any takers on this?

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Re: Probability [#permalink]

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New post 18 Jul 2008, 12:44
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young_gun wrote:
young_gun wrote:
what about assuming, without replacement?
16/35?


any takers on this?


I believe you have it right!

[RRRB + RRBR + RBRR + BRRR] + [BBBR + BBRB + BRBB + RBBB]
4 * [4/8 * 3/7 * 2/6 * 4/5] + 4*[...]
4* 2/35 + 4*2/35 = 16/35

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Re: Probability [#permalink]

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New post 18 Jul 2008, 21:56
Can this be solved using combinatorial approach ?

Total number of ways in which 4 out of 8 hats can be selected is 8C4..... and so on ?

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Re: Probability [#permalink]

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New post 19 Jul 2008, 01:17
Hi Việt,

Possibility = \(\frac{2*C_4^3 * C_4^1}{C_8^4}\) = \(\frac{16}{35}\)

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Re: Probability [#permalink]

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New post 19 Jul 2008, 10:27
lexis wrote:
Hi Việt,

Possibility = \(\frac{2*C_4^3 * C_4^1}{C_8^4}\) = \(\frac{16}{35}\)


I am not sure if combination would work here since the hats are being replaced. IMO 1/8 is the answer.

Could anyone post the OA?
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Re: Probability [#permalink]

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New post 16 Aug 2008, 21:21
I am so sorry i forgot this interesting topic.
The correct answer is 1/2
Explaination :
If we need 3 red and 1 blue:
The prob must be 4C3 * (1/2)^4
Getting three red out of 4 that are taken out has 4 options (4!/(3!*1!)) each option has a probability of (1/2)^4 since drawing a red or blue has a 50% chance.
( I saw some guys list the options as RRRB, RRBR, RBRR, BRRR; that is another way of saying 4C3 options)
The same with 3 blue and 1 red.
So the probability must be
4*(1/2)^4*2 = 1/2

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Re: Probability [#permalink]

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New post 17 Aug 2008, 23:56
i'm in deep trouble. I can't understand the equations. Is there a good tutorial or link for probabilities?

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Re: Probability [#permalink]

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New post 24 Aug 2008, 18:35
is it 25%?

4C3*(0.5)^3*(0.5)^1

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Re: Probability   [#permalink] 24 Aug 2008, 18:35
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