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

My approach:

Required Arrangement -- WRRW
- 2C1 * 2C1 * 1 * 1
- 4
Total arrangement -- 4! - 24.
Hence probability - 4/24 -- 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
Megha

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

Final Answer:

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

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

APPROACH #1: 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
Answer: B


APPROACH #2: 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
Answer: B

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?

thanks in advance

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

Hi Bunuel, I got how you got 4!/2!*2! = 6 . But then how did we infer that the probability will be 1/6?

Posted from my mobile device

There are \(\frac{4!}{2!2!}=6\) different ways to arrange W, R, R, and W:

WWRR;
WRWR;
WRRW;
RRWW;
RWRW;
RWWR.

We need the probability of WRRW case: P = favorable/total = 1/6.

Hope it's clear.

Shouldn't it be mentioned in the question that both red and both white roses are identical.

gag1994
Do they have to be identical? What if one of the red rosebushes has 1 flower and the other red rosebush has 63 flowers? Does that impact your approach to the question?
We are given only two categories of "items": red rosebushes and white rosebushes; the items within each category don't need to identical.

As others have point out (I like Brent's post best), there are different methods available. I'd attack this one as follows:
First plant has to be white. Probability 2/4 = 1/2.
Second plant has to be red. Probability 2/3.
Third plant has to be red. Probability 1/2.
Fourth plant has to be white. Probability 1/1.

1/2 * 2/3 * 1/2 * 1/1 = 2/12 = 1/6

Answer choice B.

Hi Bunel, how do we know that the two red rosebushes are identical and the two white rosebushes are identical since it is not explicitly stated in the question? Thanks for all your help on this platform, it really helps - kudos!

Because the question asks for the 2 red rosebushes to be in the middle, and nowhere does it say the gardener treats the bushes as distinct, it's clear from context that the bushes of the same color are identical.

why cant we sue the arranging formula n!/(n-k)! for the numerator?