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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]

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12 Apr 2006, 20:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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) Â¼
(c) Â½
(d) 3/8
(e) 7/12
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Re: prob - red hats, blue hats [#permalink]

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12 Apr 2006, 21:31
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) Â¼
(c) Â½
(d) 3/8
(e) 7/12

I got

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

= 16/35

none of the answers, i might be wrong
VP
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Re: prob - red hats, blue hats [#permalink]

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12 Apr 2006, 21:42
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) Â¼
(c) Â½
(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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13 Apr 2006, 00:55
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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13 Apr 2006, 01:14
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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13 Apr 2006, 03:35
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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13 Apr 2006, 09:34
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, 07:38, edited 1 time in total.
VP
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13 Apr 2006, 15:29
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

Senior Manager
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13 Apr 2006, 21:11
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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14 Apr 2006, 10:54
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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14 Apr 2006, 13:14
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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15 Apr 2006, 09:42
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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19 Apr 2006, 00:13
What's the OA?
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19 Apr 2006, 10:37
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.
_________________

-al

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19 Apr 2006, 10:50
Getting 16/35 as well
19 Apr 2006, 10:50
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# A drawer holds 4 red hats and 4 blue hats. What is the

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