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

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Manager
Joined: 12 Oct 2008
Posts: 104
A drawer holds 4 red hats and 4 blue hats. What is the [#permalink]

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14 Nov 2008, 16:17
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36. 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 immediately returning every hat to the drawer before taking out the next?

(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12

Manager
Joined: 23 Jul 2008
Posts: 193

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14 Nov 2008, 16:24
There are 4 red and 4 blue hats. Interesting to note that replacement is allowed.
Hence probability of taking out either red or blue hat will be 4/8=1/2

case one: 3 red 1 blue= 1/2^4
case two : 3 r 1 b = 1/2^4
adding the two we get probability= 1/8
Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown

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15 Nov 2008, 04:52
hibloom wrote:
There are 4 red and 4 blue hats. Interesting to note that replacement is allowed.
Hence probability of taking out either red or blue hat will be 4/8=1/2

case one: 3 red 1 blue = 1/2^4
case two : 3 b 1 r = 1/2^4
adding the two we get probability= 1/8

In both the cases, you have to consider 4 different ways the hats can be taken out.
For example in case 2: BBBR, BRBB,RBBB, BBRB

So, the probablity is 4*1/16 = 1/4
So case 1 + case 2 = 1/4 + 1/4 = 1/2
Manager
Joined: 23 Jul 2008
Posts: 193

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15 Nov 2008, 10:24
LiveStronger wrote:
hibloom wrote:
There are 4 red and 4 blue hats. Interesting to note that replacement is allowed.
Hence probability of taking out either red or blue hat will be 4/8=1/2

case one: 3 red 1 blue = 1/2^4
case two : 3 b 1 r = 1/2^4
adding the two we get probability= 1/8

In both the cases, you have to consider 4 different ways the hats can be taken out.
For example in case 2: BBBR, BRBB,RBBB, BBRB

So, the probablity is 4*1/16 = 1/4
So case 1 + case 2 = 1/4 + 1/4 = 1/2

I think i got it horribly wrong. I still can t figure out. Can you explain more elaborately
Thanks
Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown

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15 Nov 2008, 10:54
1
KUDOS
Case 1 : 3 R out of 4 R and 4B

Pick first Red ball = 4/8 = 1/2
Since replacement is allowed,
Pick second Red ball = 4/8 = 1/2
Pick third Red ball = 4/8 = 1/2
Pick blue ball(questions ask for exactly 3 R, os 4th has to be Blue) = 4/8 = 1/2

So probablity is 1/2*1/2*1/2*1/2 = 1/16
But, we can pick these 4 balls in 4 different ways BRRR, RBRR, RRBR, RRRB (Here we don't have to worry about which Red ball to pick, as they are all the same)
So, total probablity = 1/16+1/16+1/16+1/16 = 4/16 = 1/4

Same with Case 2
So, case1 and case 2 together = 1/4+1/4 = 1/2

Hope that helps
Manager
Joined: 23 Jul 2008
Posts: 193

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15 Nov 2008, 10:59
Thanks it did help +1
the mistake i made was to consider only one case. Somehow i pictured that the question said first three blue and then 4th red and vice versa.
Re: probability   [#permalink] 15 Nov 2008, 10:59
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# A drawer holds 4 red hats and 4 blue hats. What is the

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