Carcass wrote:
A garden has only equal number of roses and tulips. Two flowers are selected randomly.
Quantity A |
Quantity B |
Twice the probability that both the flowers selected are roses |
Probability that one of the flowers selected is rose and the other is tulip |
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 rosesP(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 tulipP(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 greaterAnswer: B