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

Answer: B

Cheers,
Brent

We have 4 plants, so we can arrange them in 4! ways = 24 ways

Hey, why is it 4! line all plants are different?
You have 2 white and 2 red. I assumed that you cannot distinguish between them and I counted the total amount as:
(4!)/(2!2!)

Why did you count it as you have 4 different plants?

We are asked to find the probability of one particular pattern: WRRW.

Total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters WWRR, out of which 2 W's and 2 R's are identical, so \(\frac{4!}{2!2!}=6\);

So \(p=\frac{1}{6}\).

Answer: B.


Or you can do with direct probability approach:

The probability that the first rosebush will be white is 2/4 (there are two white out of total 4 rosebushes);
The probability that the second rosebush will be red is 2/3 (there are two reds out of total 3 rosebushes left);
The probability that the third rosebush will be red is 1/2 (there are now one red out of total 2 rosebushes left);
Finally, only one white is left so the probability is 1;

P(WRRW)=2/4*2/3*1/2*1=1/6.

Answer: B.