A gardener is going to plant 2 red rosebushes and 2...

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

A gardener is going to plant 2 red rosebushes and 2...

by BTGmoderatorLU » Tue Jan 23, 2018 6:06 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 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.

I'm really confused with this PS question. Experts, any suggestion? Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Tue Jan 23, 2018 6:06 am

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by Brent@GMATPrepNow » Tue Jan 23, 2018 6:10 am

LUANDATO 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 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
= 1/6
= B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Tue Jan 23, 2018 6:11 am

LUANDATO 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 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:

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

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by regor60 » Tue Jan 23, 2018 7:47 am

Or, how many distinct permutations are there

Number of permutations: 4*3*2*1 = 24

Divide by those that are the same, in this case, two red and two white 24/2!*2! = 6 distinct permutations

Only one of these has two red in the middle, therefore [spoiler]1/6[/spoiler]

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by DavidG@VeritasPrep » Tue Jan 23, 2018 9:48 am

LUANDATO 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 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.

I'm really confused with this PS question. Experts, any suggestion? Thanks in advance.

You could also just start by selecting bushes for the middle of the row. Probability that the first bush in the middle will be red = 2/4. Probability that second bush in the middle with also be red given that the first bush was red = 1/3.

(2/4) * (1/3) = 2/12 = 1/6. The answer is B

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by [email protected] » Tue Jan 23, 2018 7:27 pm

Hi LUANDATO,

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

Contact Rich at [email protected]

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed Jan 24, 2018 9:46 am

LUANDATO 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 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

Alternate solution:

There are 4!/(2! x 2!) = 24/(2 x 2) = 6 ways (or arrangements) to plant these rosebushes. Having the two red rosebushes in the middle (i.e., white-red-red-white) is one of the 6 arrangements; thus, the probability is 1/6.

Answer: B

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Head of GMAT Instruction
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