Source: GMAT Paper Tests
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
The OA is B.
A gardener is going to plant 2 red rosebushes and 2 white
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Moving LEFT TO RIGHT along the row:BTGmoderatorLU wrote:Source: GMAT Paper Tests
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
P(1st rosebush is white) = 2/4. (Of the 4 rosebushes, 2 are white.)
P(2nd rosebush is red) = 2/3. (Of the 3 remaining rosebushes, 2 are red.)
P(3rd rosebush is red) = 1/2. (Of the 2 remaining rosebushes, 1 is red.)
P(4th rosebush is white) = 1/1. (The one remaining rosebush is white.)
Since all of these events must happen in order for the 2 middle rosebushes to be red, we MULTIPLY the fractions:
2/4 * 2/3 * 1/2 * 1 = 1/6.
The correct answer is B.
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As with many probability questions, we can also solve this using counting techniques.BTGmoderatorLU wrote:Source: GMAT Paper Tests
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
The OA is B.
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
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We can also apply probability rules:BTGmoderatorLU wrote:Source: GMAT Paper Tests
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
The OA is B.
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
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\[\left\{ \begin{gathered}BTGmoderatorLU wrote:Source: GMAT Paper Tests
A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them in a row, what is the probability that the 2 rosebushes in the middle of the row will be the red rosebushes?
A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. 1/2
\,2\,\,{\text{red}} \hfill \\
\,2\,\,{\text{white}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,P\left( {\,2\,\,{\text{red}}\,\,{\text{are}}\,\,{\text{in}}\,{\text{the}}\,\,{\text{middle}}} \right)\]
\[{\text{Total}}\,\, = \,\,4!\,\,\,{\text{equiprobable}}\,\,{\text{ways}}\]
\[{\text{Favorable}} = 2!\,\,\,\,({\text{for}}\,\,{\text{white}}\,\,1{\text{st}}\,\,{\text{and}}\,\,4{\text{th}}\,{\text{positions}})\,\,\,\,\, \cdot \,\,\,\,\,2!\,\,\,\,\,({\text{for}}\,\,{\text{red}}\,\,2{\text{nd}}\,\,{\text{and}}\,\,3{\text{rd}}\,{\text{positions}})\]
\[? = \frac{{\,2!\,\, \cdot 2!\,}}{{4!}} = \,\,\frac{1}{6}\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Hi All,
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
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
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We need to determine the probability of white-red-red-white.BTGmoderatorLU wrote:Source: GMAT Paper Tests
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
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
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