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A gardener is going to plant 2 red rosebushes

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A gardener is going to plant 2 red rosebushes

by canbtg » Thu Oct 31, 2013 3:19 am
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
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by Uva@90 » Thu Oct 31, 2013 3:29 am
canbtg 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
Hi Canbtg,
From question we get to know that there are 2 red rosebushes and 2 white rosebushes
and we want to arrange in below order

WRRW
(W=white rosebushes and R= 2 red rosebushes)

Probability of selecting First(white)one is 2/4(There are 2 white Rosebushes)
Probability of selecting Second (Red) one is 2/3(there are 2 red one out of remaining 3)
Probability of selecting third (red) one is 1/2(there is only 1 red one out of remaining 2)
Probability of selecting fourth (White) one is 1(Only one is left)

so WRRW = (2/4)*(2/3)*(1/2)*(1) = 1/6

Hence answer is B

Hope it helps you.

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by ganeshrkamath » Thu Oct 31, 2013 4:17 am
canbtg 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
Probability of selecting white rosebush first = (2/4)
Probability of selecting red rose bush second = (2/4)(2/3)
Probability of selecting red rose bush third = (2/4)(2/3)(1/2)
Probability of selecting white rose bush last = (2/4)(2/3)(1/2)(1)
= 1/6

Choose B

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by GMATGuruNY » Thu Oct 31, 2013 5:28 am
canbtg 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
Moving LEFT TO RIGHT along the row:
P(1st rosebush is white) = 2/4. (Of the 4 rosebushes, 2 are white.)
P(2nd rosebush is red) = 2/3. (Of the 3 remaining rosebushes, 2 are red.)
P(3rd rosebush is red) = 1/2. (Of the 2 remaining rosebushes, 1 is red.)
P(4th rosebush is white) = 1/1. (The one remaining rosebush is white.)
Since all of these events must happen in order for the 2 middle rosebushes to be red, we MULTIPLY the fractions:
2/4 * 2/3 * 1/2 * 1 = 1/6.

The correct answer is B.
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hi

by Jeff@TargetTestPrep » Thu Dec 07, 2017 7:30 am
canbtg 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
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

Answer: B

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by Brent@GMATPrepNow » Thu Dec 07, 2017 7:37 am
canbtg 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
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)
= [spoiler]1/6[/spoiler]
= B

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by Brent@GMATPrepNow » Thu Dec 07, 2017 7:38 am
canbtg 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
We can also apply probability rules here.
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
= [spoiler]1/6[/spoiler]
= B

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by [email protected] » Thu Dec 07, 2017 12:59 pm
Hi All,

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

Final Answer: B

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by Matt@VeritasPrep » Thu Dec 07, 2017 4:26 pm
Another resurrected zombie thread - what is happening in the graveyard this week? :)
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