A gardener is going to plant 2 red redbushes and 2 white redbushes.If the gardener is to select each of the bushes at random,one at a time and plant them in a row, what is the probability that the 2 redbushes in the middle of the row will be redbushes.
a) 1/12
b) 1/6
c) 1/5
d) 1/3
e) 1/2
Gardener
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The total number of ways the four bushes will be arranged = 4! = 24
Now we have W R R W as the required arrangement.
The 2 red bushes can be arranged amongst themselves in 2 ways. Same holds good for the white bushes as well.
Hence required probability = (2* 2) / 24 = 1/6
Now we have W R R W as the required arrangement.
The 2 red bushes can be arranged amongst themselves in 2 ways. Same holds good for the white bushes as well.
Hence required probability = (2* 2) / 24 = 1/6
- gmat740
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Can you tell me where I am wrong :
Select one red from 2 => 2C1 = 2
Select second red from 1 left red redbushes =>2C2 = 1
Total ways of selection = 2*1 = 2
Now these can be arrange in 2! ways, so total ways = 2*2! = 4
similarly, the other 2 white redbushes can arrange themselves in 2! ways
thus total ways = 4*2= 8
Prob = 8/24 = 1/3
Where am I wrong? Aren't we going to make a selection of 2 redbushes from the entire set??
How did you directly arrange without selecting?
Select one red from 2 => 2C1 = 2
Select second red from 1 left red redbushes =>2C2 = 1
Total ways of selection = 2*1 = 2
Now these can be arrange in 2! ways, so total ways = 2*2! = 4
similarly, the other 2 white redbushes can arrange themselves in 2! ways
thus total ways = 4*2= 8
Prob = 8/24 = 1/3
Where am I wrong? Aren't we going to make a selection of 2 redbushes from the entire set??
How did you directly arrange without selecting?
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You're double counting selection and arranging, even though you mean the same thing.gmat740 wrote:Can you tell me where I am wrong :
Select one red from 2 => 2C1 = 2
Select second red from 1 left red redbushes =>2C2 = 1
Total ways of selection = 2*1 = 2
Now these can be arrange in 2! ways, so total ways = 2*2! = 4
similarly, the other 2 white redbushes can arrange themselves in 2! ways
thus total ways = 4*2= 8
Prob = 8/24 = 1/3
Where am I wrong? Aren't we going to make a selection of 2 redbushes from the entire set??
How did you directly arrange without selecting?
Let's just think this one out logically:
We want W R R W
There are 2 possible white bushes for first slot, 2 red bushes for 2nd slot, 1 red for 3rd slot (once we've placed one red bush) and 1 white for 4th slot (once we've placed 1st white bush), so:
2*2*1*1 = 4 desired arrangements.
There are 4! = 24 total possible arrangements, so our answer is 4/24 = 1/6
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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