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.

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?

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

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.

= \(\frac{1}{6}\)

Answer B


ALL THE BEST

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

Duplicate response.
Deleted.

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 \(\frac{1}{6}\)

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

The method i used was to consider the 2 red rosebushes as one.
Hence WRW
Three objects can be arranged in 3!=6 ways.
only one possible arrangement i.e. WRW
p= 1/6

This is a permutation question.

4! / 2! x 2! = 6 <--- # of ways you can arrange 2 red and 2 white rosebushes

Out of those 6 ways ONLY 1 way arrangement is W R R W

1/6

B.

When we find total number of ways i.e 24 (4!), wouldn't we divide it by 2!, 2! for identical cases ?

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.