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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

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Source: GMAT Paper Tests

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

The OA is B.

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BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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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BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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

The OA is B.
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
= B

Cheers,
Brent

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Brent Hanneson – Creator of GMATPrepNow.com
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BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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

The OA is B.
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
= B

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

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Post
BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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
\[\left\{ \begin{gathered}
\,2\,\,{\text{red}} \hfill \\
\,2\,\,{\text{white}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,P\left( {\,2\,\,{\text{red}}\,\,{\text{are}}\,\,{\text{in}}\,{\text{the}}\,\,{\text{middle}}} \right)\]
\[{\text{Total}}\,\, = \,\,4!\,\,\,{\text{equiprobable}}\,\,{\text{ways}}\]
\[{\text{Favorable}} = 2!\,\,\,\,({\text{for}}\,\,{\text{white}}\,\,1{\text{st}}\,\,{\text{and}}\,\,4{\text{th}}\,{\text{positions}})\,\,\,\,\, \cdot \,\,\,\,\,2!\,\,\,\,\,({\text{for}}\,\,{\text{red}}\,\,2{\text{nd}}\,\,{\text{and}}\,\,3{\text{rd}}\,{\text{positions}})\]
\[? = \frac{{\,2!\,\, \cdot 2!\,}}{{4!}} = \,\,\frac{1}{6}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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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

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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