Problem Solving

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

I will go like this :

Total no of ways 4 bushes can be arranged:

4! /(2!*2!) (since two red and two balck bushes are same..)

No of ways red are in middle: 1 BRRB (since you cannot distinguis between 2 balck or 2 red bushes..)

so probablity= 1/6.

Not sure I am wrong ..

What is OA?

smallsorrow 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

smallsorrow 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

Red bushes can be arranged in four ways.

W1R1R2W2
W2R1R2W1
W1R2R1W2
W2R2R1W1

total no. of ways 4 bushes can be arranged = 4! = 24

probability = 4/24 = 1/6

OA?

1/6

In this Question,
Whether you ASSUME both red rosebushes and blue rose bushes are identical...
the answer is the same...1/6

Assume red are identical , blue are identical

no of the desired arrangements..brrb = 1

no of. arrangements ...4!/2!*2! =6
=> p=1/6

Assume they are not identical

no. of desired arrangements..[b1r1r2b2, b1r2r1b2, b2r1r2b1, b2r2r1b1] =4
no of . arrangements =4! =24

=> p=4/24 =1/6

I ASSUMED THAT THEY ARE IDENTICAL AND SOLVED THE PROBLEM.
HOWEVER THE ANSWER IS THE SAME IN BOTH APPROACHES.IT MIGHT NOT HAPPEN THIS WAY IN OTHER PROBLEMS. :roll:

COULD ANY OF THE INSTRUCTORS PLEASE HELP ME WHICH IS THE RIGHT WAY ??

THANKS A TON!!

Assume red are identical , blue are identical

Until and unless nothing has been told about the rosebushes specifically, we cannot assume them to be different

Rashmi1804 wrote:In this Question,
Whether you ASSUME both red rosebushes and blue rose bushes are identical...
the answer is the same...1/6

Assume red are identical , blue are identical

no of the desired arrangements..brrb = 1

no of. arrangements ...4!/2!*2! =6
=> p=1/6

Assume they are not identical

no. of desired arrangements..[b1r1r2b2, b1r2r1b2, b2r1r2b1, b2r2r1b1] =4
no of . arrangements =4! =24

=> p=4/24 =1/6

I ASSUMED THAT THEY ARE IDENTICAL AND SOLVED THE PROBLEM.
HOWEVER THE ANSWER IS THE SAME IN BOTH APPROACHES.IT MIGHT NOT HAPPEN THIS WAY IN OTHER PROBLEMS. :roll:

COULD ANY OF THE INSTRUCTORS PLEASE HELP ME WHICH IS THE RIGHT WAY ??

THANKS A TON!!

Inviting instructor's comments here please...!!

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


I've never posted on a GMAT forum before, but I have taught the GMAT for 10 years...

I have developed a 5-step method for probability of multiple events which can be applied as follows:
1) Lay out the number of events (here, we have 4 events):
_ _ _ _

2) Label the events with one specific example of the desired outcome (here, we want 2 reds in the middle):
_ _ _ _
w1 r1 r2 w2

3) Assign the relevant probability to each event and multiply across (here, we have a "selection" situation where elements are removed):

1/4 1/3 1/2 1 = 1/24
w1 r1 r2 w2

(NOTE: the probability of selecting w2 once every other element has been selected is 100%. This is a concept somewhat separate from the method I'm discussing now).

(NOTE: the resulting product is known as the "specific probability" (the probability of one specific event).

4) Determine the number of ways in which the desired outcome can be presented (here, we have 4 ways):
w1 r1 r2 w2
w2 r1 r2 w1
w2 r2 r1 w1
w1 r2 r1 w2

(NOTE: This step often requires "counting methods", such as permutation or combination).

5) Multiply the result of step 3 (the specific probability) by the result of step 4 (the number of ways):

1/24 x 4 = 4/24 = 1/6

(NOTE: The specific probabilities of each outcome could actually be different in some circumstances. In this case, add the specific probabilities together).

(NOTE: For "at least" questions, use the same method to find the probability of the desired event NOT occurring and subtract the result from 1).

(NOTE: When dealing with large numbers, using the more traditional methods described by others works better, but you would still have to deal with combination and permutation methods)

smallsorrow 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 solve using simple probability.

Probability = # of desired outcomes / total # of possibilities

When we want the probability of MULTIPLE events, we MULTIPLY the individual probabilities.

Prob 1st bush white: 2/4
Prob 2nd bush red (after first white): 2/3
Prob 3rd bush red (after first white/second red): 1/2
Prob 4th bush white (after white/red/red): 1/1

So, probability of getting WRRW: (2/4)(2/3)(1/2)(1/1) = 4/24 = 1/6

We really don't care if the bushes are identical, since all that matters is colour. Even though this question seems like complicated combinatorics, it's identical to:

There are 4 socks in a drawer, 2 white and 2 red. If you select the socks one at a time, without replacement, what's the probability that the order is WRRW?