Let's assign some variables...
Let \(r\) = the number of roses in the garden
Let \(\)t = the number of tulips in the garden
Since garden has only equal number of roses and tulips, we know that \(r = t\)

QUANTITY A: Twice the probability that both the flowers selected are roses
P(both flowers are roses) = P(1st flower is a rose AND 2nd flower is a rose)
= P(1st flower is a rose) x P(2nd flower is a rose)

= \(\frac{r}{r+t}\) x \(\frac{r-1}{r+t-1}\)

= \(\frac{r(r-1)}{(r+t)(r+t-1)}\)

We want TWICE that probability, which means QUANTITY A = \(\frac{2r(r-1)}{(r+t)(r+t-1)}\)


QUANTITY B: Probability that one of the flowers selected is rose and the other is tulip
P(select one rose and one tulip) = P(1st flower is a rose AND 2nd flower is a tulip OR 1st flower is a tulip AND 2nd flower is a rose)
= [P(1st flower is a rose) x P(2nd flower is a tulip)] + [P(1st flower is a tulip) x P(2nd flower is a rose)]

= [\(\frac{r}{r+t}\) x\(\frac{t}{r+t-1}\)] + [\(\frac{t}{r+t}\) x \(\frac{r}{r+t-1}\)]

= \(\frac{rt}{(r+t)(r+t-1)}\) + \(\frac{tr}{(r+t)(r+t-1)}\)

= \(\frac{2rt}{(r+t)(r+t-1)}\)

So we have the following:
QUANTITY A: \(\frac{2r(r-1)}{(r+t)(r+t-1)}\)

QUANTITY B: \(\frac{2rt}{(r+t)(r+t-1)}\)

Multiply both quantities by \((r+t)(r+t-1)\) to get:
QUANTITY A: \(2r(r-1)\)
QUANTITY B: \(2rt\)

Divided both quantities by \(2r\) to get:
QUANTITY A: \(r-1\)
QUANTITY B: \(t\)

Since \(r = t\), we can rewrite Quantity B as follows:
QUANTITY A: \(r-1\)
QUANTITY B: \(r\)

At this point we can clearly see that Quantity B is greater

Answer: B