BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:42 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 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

Answer: B
Source: GMAT Paper Tests

Let's 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

Answer: B

Cheers,

BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:42 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 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

Answer: B
Source: GMAT Paper Tests

Solution:

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:

Using the indistinguishable permutations formula, we see that 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