Problem Solving

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

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

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

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