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# A gardener is going to plant 2 red rosebushes and 2 white

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A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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26 Aug 2010, 07:38
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Question Stats:

69% (00:59) correct 31% (01:03) wrong based on 508 sessions

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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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½
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A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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26 Aug 2010, 07:55
14
18
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

We are asked to find the probability of one particular pattern: WRRW.

Total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters WWRR, out of which 2 W's and 2 R's are identical, so $$\frac{4!}{2!2!}=6$$;

So $$p=\frac{1}{6}$$.

Or you can do with direct probability approach:

The probability that the first rosebush will be white is 2/4 (there are two white out of total 4 rosebushes);
The probability that the second rosebush will be red is 2/3 (there are two reds out of total 3 rosebushes left);
The probability that the third rosebush will be red is 1/2 (there are now one red out of total 2 rosebushes left);
Finally, only one white is left so the probability is 1;

P(WRRW)=2/4*2/3*1/2*1=1/6.

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26 Aug 2010, 19:42
4
My approach:

Required Arrangement -- WRRW
- 2C1 * 2C1 * 1 * 1
- 4
Total arrangement -- 4! - 24.
Hence probability - 4/24 -- 1/6.
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10 Jan 2012, 01:08
5
M8 wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. 1/2

You can also look at the probability of each intermediate step.

Probability that he first plants a white one = 2/4 (There are 2 whites and 2 reds)
Probability that he then plants a red one = 2/3 (There are 1 white and 2 reds)
Probability that he again plants a red one = 1/2 (There are 1 white and 1 red)
Now only the white one is left so he plants it.
Total probability = (2/4)*(2/3)*(1/2)*1 = 1/6
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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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17 Feb 2016, 16:04
Bunuel wrote:
Bumping for review and further discussion.

Hello Bunuel, question for you, or anyone else who can confirm and clarify.

According to probability rules, we start with the most restrictive stages, and work our way from there.

The question asks what is the probably that the two red rosebushes in the middle of the row will be red, which gives us the restriction which can be interpreted as the most strict being that 2 in the middle are Red or the 2 on the outside are white, basically being the same.

Spot 1 Spot 2 Spot 3 Spot 4

Spot 1 - Has to be white, thus is 2/4 1/2
Spot 4 - (Continuing with the restriction that both outsides have to be white) is 1/3
Spot 2 - Has to be red, and since we have two red rosebushes left, it is 1
Spot 3 - Is also 1 since only the red rosebush is left

1/2 x 1 x 1 x 1/3

Multiplied together give us the answer of 1/6

Is there a mistake in my way of thinking?
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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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11 Sep 2016, 13:31
Bunuel wrote:
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

We are asked to find the probability of one particular pattern: WRRW.

Total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters WWRR, out of which 2 W's and 2 R's are identical, so $$\frac{4!}{2!2!}=6$$;

So $$p=\frac{1}{6}$$.

Or you can do with direct probability approach:
The probability that the first rosebush will be white is 2/4 (there are two white out of total 4 rosebushes);
The probability that the second rosebush will be red is 2/3 (there are two reds out of total 3 rosebushes left);
The probability that the third rosebush will be red is 1/2 (there are now one red out of total 2 rosebushes left);
Finally, only one white is left so the probability is 1;

P(WRRW)=2/4*2/3*1/2*1=1/6.

This 1 particular pattern WRRW , can it not be arranged in 4 ways , 2 ways that white bushes can be interchanged
and 2 ways that red bushes can be interchanged , so total 4 ways and as per my understanding shouldn't answer be 4/6.
Would really appreciate your help here.

Thanks
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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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11 Sep 2016, 18:52
Hi Megha,

The math that's required to answer this question can actually be done in a couple of different ways, depending on how you "see" probability questions.

We're given 2 red rosebushes (R1 and R2) and two white rosebushes (W1 and W2). We're told to put these 4 rosebushes in a row; the question asks for the probability that the "middle two" rosebushes are both red.

Probability is defined as…

(# of ways that you want)/(# of ways that are possible)

The # of ways that are possible = (4)(3)(2)(1) = 24 possible ways to arrange the 4 bushes.

The specific ways that we want have to fit the following pattern:

W-R-R-W

The first bush must be white; there are 2 whites
The second bush must be red; there are 2 reds
The third bush must be red, but after placing the first red bush, there's just 1 red left
The fourth bush must be white, but after placing the first white bush, there's just 1 white left

= (2)(2)(1)(1) = 4

4 ways that fit what we want
24 ways that are possible

4/24 = 1/6

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A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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08 Nov 2016, 03:52
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

To Find the probability of W R R W

P(W) * P(R) * P(R) * P(W)

$$\frac{1}{4} * \frac{1}{3} * \frac{1}{2} * 1 * 2! * 2!$$

2! * 2! for number of ways 2 RED and 2 WHITE can change their positions.

= $$\frac{1}{6}$$

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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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09 Nov 2016, 11:00
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

We need to determine the probability of white-red-red-white.

Let’s determine the probability of each selection.

1st selection:

P(white rosebush) = 2/4 = 1/2

2nd selection:

P(red rosebush) = 2/3

3rd selection:

P(red rosebush) = 1/2

4th selection:

P(white rosebush) =1/1 = 1

Thus, P(white-red-red-white) =1/2 x 2/3 x 1/2 x 1 = 1/6

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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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23 Jan 2018, 07:41
Top Contributor
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

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,
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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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23 Jan 2018, 07:42
Top Contributor
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

We can also apply probability rules:

P(2 middle bushes are red) = P(1st bush is white AND 2nd bush is red AND 3rd bush is red AND 4th bush is white)
= P(1st bush is white) x P(2nd bush is red) x P(3rd bush is red) x P(4th bush is white)
= 2/4 x 2/3 x 1/2 x 1/1
= 1/6

Cheers,
Brent
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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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23 Jan 2018, 09:53
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

Though i know these combinatircs formula and the probability rules, i couldnt apply a correct technique under time constraints

so i solved it like this

WRRW
WRWR
RWRW
WWRR
RRWW
RWWR

total number of possibilities 6. hence $$\frac{1}{6}$$
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A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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24 Jan 2018, 06:03
GMATPrepNow wrote:
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

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,

Dear GMATPrepNow

Why do we consider that there are 2 different types of white or bushes? Is not R1 the same as R2 ( and also W1 & W2) so that (W1, R1, R2, W2) should be same as (W1, R2, R1, W2) and (W2, R1, R2, W1) and ( W2, R2, R1, W1)? All should treated as one arrangement.

Based on my understanding, I did it as follows:

W W R R
W R W R
W R R W
R R W W
R W R W
R W W R

P(2 middle are red) = 1/6..........Same as you got.

Did I go wrong in my solution above?

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Re: A gardener is going to plant 2 red rosebushes and 2 white  [#permalink]

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24 Jan 2018, 15:28
Top Contributor
Mo2men wrote:
GMATPrepNow wrote:
udaymathapati wrote:
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 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. ½

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,

Dear GMATPrepNow

Why do we consider that there are 2 different types of white or bushes? Is not R1 the same as R2 ( and also W1 & W2) so that (W1, R1, R2, W2) should be same as (W1, R2, R1, W2) and (W2, R1, R2, W1) and ( W2, R2, R1, W1)? All should treated as one arrangement.

Based on my understanding, I did it as follows:

W W R R
W R W R
W R R W
R R W W
R W R W
R W W R

P(2 middle are red) = 1/6..........Same as you got.

Did I go wrong in my solution above?

Your solution and my solution are both valid.
In your solution, you treated the same-colored bushes as the same and, more importantly, you applied this to BOTH numerator and denominator.
In my solution, I treated the same-colored bushes as different, and, more importantly, I applied this to BOTH numerator and denominator.

Cheers,
Brent
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Re: A gardener is going to plant 2 red rosebushes and 2 white &nbs [#permalink] 24 Jan 2018, 15:28
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