Probablity/P&C problem Exam Pack1

Question

Please see image attached

GMATGuruNY · Answer

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

Brent Hanneson - Creator of GMATPrepNow.com

Rich.C@EMPOWERgmat.com · Answer

Hi sukhman,

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

Contact Rich at [email protected]

Brent@GMATPrepNow · Answer

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

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

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

Brent Hanneson - Creator of GMATPrepNow.com

hoppycat · Answer

Rich.C@EMPOWERgmat.com wrote:Hi sukhman, 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 Can you explain the part in red.

Matt@VeritasPrep · Answer

hoppycat wrote: Can you explain the part in red.

Typically this means either finding all the arrangements, then finding the number of arrangements that do what you want OR thinking step by step about what needs to go where.

If we try it the first way (arrangements), we have RRWW in some order. There are 4!/2!2! = 6 ways to arrange those, of which only one (WRRW) gives us what we want.

If we try it the second way (step by step), we could say:

The first bush MUST be white. There are 2 whites out of 4 total, so this is 2/4.
The second bush MUST be red. There are 2 reds out of 3 bushes left, so this is 2/3.
The third bush MUST be red. There is 1 reds out of 2 bushes left, so this is 1/2.
The fourth bush is determined now, so we're done: 2/4 * 2/3 * 1/2 = 4/24 = 1/6.

Matt@VeritasPrep · Answer

Footnote: it's definitely a good idea to practice as many probability questions as you can both ways: arranging and/or analyzing. Most problems that can be solved either way have one approach that is quite a bit simpler and easier than the other.

Scott@TargetTestPrep · Answer

sukhman wrote: ↑Sat Jan 23, 2016 2:28 am Please see image attached Solution: We need to determine the probability of white-red-red-white. 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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