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Probability - HELP

by soni_pallavi » Mon Oct 29, 2012 2:51 am
Hi

Could anyone show the working of this question.

A gardener is going to plant 2 red rosebushes and 2 white rosebushes,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 two rose bushes in the middle will be red??

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by Anurag@Gurome » Mon Oct 29, 2012 4:00 am
soni_pallavi wrote:Hi

Could anyone show the working of this question.

A gardener is going to plant 2 red rosebushes and 2 white rosebushes,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 two rose bushes in the middle will be red??

Thanks
Probability that the first rosebush will be white = 2/4
Now there are 2 reds out of total 3 rosebushes left, so probability that the second rosebush will be red = 2/3
Now one red rosebush out of total 2 rosebushes is left, so probability that the third rosebush will be red = 1/2
Now, one white rosebush is left so the probability = 1

Probability that 2 rose bushes in the middle will be red = 2/4 * 2/3 * 1/2 * 1 = [spoiler]1/6[/spoiler]
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by Brent@GMATPrepNow » Mon Oct 29, 2012 5:13 am
soni_pallavi wrote:Hi

Could anyone show the working of this question.

A gardener is going to plant 2 red rosebushes and 2 white rosebushes,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 two rose bushes in the middle will be red??

Thanks
As with many probability questions, we can also solve this using counting techniques.

P(2 middle are red) = (# of outcomes with 2 red in middle)/(total number of outcomes)

Label the 4 bushes as W1, W2, R1, R2

total number of outcomes
We have 4 plants, so we can arrange them in 4! ways = 24 ways

# of outcomes with 2 red in middle
If we consider the possibilities here, we can list them very quickly:
- W1, R1, R2, W2
- W1, R2, R1, W2
- W2, R1, R2, W1
- W2, R2, R1, W1
So, there are 4 outcomes with 2 red in middle

P(2 middle are red) = (4)/(24)
= 1/6

Cheers,
Brent
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by gmattesttaker2 » Mon May 19, 2014 7:42 am
Brent@GMATPrepNow wrote:
soni_pallavi wrote:Hi

Could anyone show the working of this question.

A gardener is going to plant 2 red rosebushes and 2 white rosebushes,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 two rose bushes in the middle will be red??

Thanks
As with many probability questions, we can also solve this using counting techniques.

P(2 middle are red) = (# of outcomes with 2 red in middle)/(total number of outcomes)

Label the 4 bushes as W1, W2, R1, R2

total number of outcomes
We have 4 plants, so we can arrange them in 4! ways = 24 ways

# of outcomes with 2 red in middle
If we consider the possibilities here, we can list them very quickly:
- W1, R1, R2, W2
- W1, R2, R1, W2
- W2, R1, R2, W1
- W2, R2, R1, W1
So, there are 4 outcomes with 2 red in middle

P(2 middle are red) = (4)/(24)
= 1/6

Cheers,
Brent
Hello Brent,

I was just wondering if the following approach is also OK:

I first select two red rose bushes for the middle i.e.

- 2/4 1/3 -

I then select the two white bushes for the sides i.e.

2/2 2/4 1/3 1/1

= 1/6

Thanks a lot for your help.

Best Regards,
Sri
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by Brent@GMATPrepNow » Mon May 19, 2014 8:24 am
That solution is perfect, Sri.
Great approach!!

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by [email protected] » Mon May 19, 2014 10:52 am
Hi Sri,

You can use your approach and work "from left to right"; you'll get the same answer:

We want W R R W

2/4 2/3 1/2 1/1

= 4/24 = 1/6

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by parveen110 » Mon Jun 02, 2014 2:15 am
Just would like to implement one of the important things about probablity that i learnt through many posts on this forum: It doesn't matter how you count things, as long as you're consistent.

Coming to the problem:
Let two red and two white rose bushes are identical.

So, The desired combination is: WRRW

Total outcomes= 4!/2!.2!=24/4=6

Favourable outcome= 1

Probablity of planting two red bushes in the middle= FO/TO= 1/6
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