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

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A drawer holds 4 red hats and 4 blue hats. What is the [#permalink] New post 12 Apr 2006, 19:18
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?
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12
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Re: prob - red hats, blue hats [#permalink] New post 12 Apr 2006, 20:31
chillpill 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?
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12



I got

2 * ((4c3)*(4c1)) / (8c4)

= 16/35

none of the answers, i might be wrong
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Re: prob - red hats, blue hats [#permalink] New post 12 Apr 2006, 20:42
conocieur wrote:
chillpill 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?
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12

I got 2 * ((4c3)*(4c1)) / (8c4) = 16/35.
none of the answers, i might be wrong


absolutly agree. thats what i got too.

= 2 [(4c3)*(4c1)] / (8c4) = 32/70 = 16/35
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 [#permalink] New post 12 Apr 2006, 23:55
If we use binominal formula (0,5+0,5)^4 the prob of getting 3R1B or 3B1R is 3C4*0,5*0,5^3=4/16=1/4
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 [#permalink] New post 13 Apr 2006, 00:14
BG wrote:
If we use binominal formula (0,5+0,5)^4 the prob of getting 3R1B or 3B1R is 3C4*0,5*0,5^3=4/16=1/4


can you explain this one? i was getting 16/35 myself too
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 [#permalink] New post 13 Apr 2006, 02:35
I am also getting 16/35

Total outcomes = C(8,4)

Favourable = C(4,3) * C (4,1) * 2

Probability = (C(4,3) * C(4,1) * 2 ) / C(8,4) = 16/35
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 [#permalink] New post 13 Apr 2006, 08:34
Made a mistake, i agree with 16/35.

Since it is "OR" should we not be adding.

Here is my take, i might be wrong.

Denominator= 8C4= 70

Numerator,

3 red hats out of 4 red hats and 1 Blue from 4 blue= 4C3*4C1=16
3 Blue hats out of 4 blue hats and 1 red from 4 red= 4C3*4C1=16

16/70 + 16/70= 16/35

Still i don;t see this in the answer choice.

Thanks

Last edited by swaroh on 14 Apr 2006, 06:38, edited 1 time in total.
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 [#permalink] New post 13 Apr 2006, 14:29
BG wrote:
If we use binominal formula (0,5+0,5)^4 the prob of getting 3R1B or 3B1R is 3C4*0,5*0,5^3=4/16=1/4


BG can you please elaborate?
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 [#permalink] New post 13 Apr 2006, 20:11
Professor would you please elaborate the numerator
2 [(4c3)*(4c1)]?
I see wht is 4C3, but what is '2' and 4C1 and why you multiply them?

Thank you.
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 [#permalink] New post 14 Apr 2006, 09:54
BG wrote:
If we use binominal formula (0,5+0,5)^4 the prob of getting 3R1B or 3B1R is 3C4*0,5*0,5^3=4/16=1/4



I thought about this approach too, even though it doesn't make too much sense, cause it would imply that every time you get one hat you put it back in the drawer and then pick a hat again, I would assume there is not replacement, but anyway even if we take BG's approach the answer would not be 1/4 but 1/2 as we have to consider 3reds - 1 blue or 1 red - 3 blues.

so taking the options from the question it would be

1/2
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 [#permalink] New post 14 Apr 2006, 12:14
1/2 indeed

Bionomial situation

3 R = 4C3 x 1/16

3R or 3R = 2(4C3 x 1/16) = 1/2
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 [#permalink] New post 15 Apr 2006, 08:42
Unless I didn't understand the question properly, my answer is 2/5.

chillpill, pls post the OA and OE. And the source of the question as well.

Thanks,
Vipin
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 [#permalink] New post 18 Apr 2006, 23:13
What's the OA? :)
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 [#permalink] New post 19 Apr 2006, 09:37
Could it be 1/8 ?:)

Let p be the prob. to choose a red hat, and let q be the prob. to choose a blue one.

We should find P(exactly 3 blue or exactly 3 red)=qp^3+pq^3, where p = 4/8 and q = 4/8.
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-al

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 [#permalink] New post 19 Apr 2006, 09:50
Getting 16/35 as well
  [#permalink] 19 Apr 2006, 09:50
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