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Probability

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by RiyaR » Thu Oct 09, 2014 10:01 pm
A gardner is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the rosebushes at random, one at a time, and plant them in a row, what is the probability that the two red rosebushes in the middle of the row will be red?

A)1/12
B)1/6
C)1/5
D)1/3
E)1/2
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by [email protected] » Thu Oct 09, 2014 10:30 pm
Hi RiyaR,

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

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by Brent@GMATPrepNow » Fri Oct 10, 2014 4:44 am
soni_pallavi wrote: 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 two rose bushes in the middle will be red??

A)1/12
B)1/6
C)1/5
D)1/3
E)1/2
As with many probability questions, we can also 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)
= [spoiler]1/6[/spoiler]
= B

Cheers,
Brent
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by Brent@GMATPrepNow » Fri Oct 10, 2014 4:49 am
RiyaR wrote:A gardner is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the rosebushes at random, one at a time, and plant them in a row, what is the probability that the two red rosebushes in the middle of the row will be red?

A)1/12
B)1/6
C)1/5
D)1/3
E)1/2
We can also apply probability rules here.
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
= [spoiler]1/6[/spoiler]
= B

Cheers,
Brent
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by Matt@VeritasPrep » Sat Oct 11, 2014 10:29 am
Suppose the bushes are W, W, R, and R. (It won't matter if they're unique, so we'll treat the two white ones as identical and the two red ones as identical.)

We can arrange them in six ways:

WWRR
RRWW
WRWR
RWRW
WRRW
RWWR

Only the highlighted arrangement works, so only 1/6 of our arrangements are acceptable.
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