BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probablity/P&C problem Exam Pack1

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Jan 23, 2016 4:12 am
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
Moving LEFT TO RIGHT along the row:
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Sat Jan 23, 2016 10:25 am
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]
Image
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 23, 2016 10:36 am
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
Image
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 23, 2016 10:37 am
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
Image
Top
Senior | Next Rank: 100 Posts
Posts: 36
Joined: Sun May 21, 2017 5:41 am

by hoppycat » Wed May 31, 2017 5:01 am
[email protected] 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.
Top
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Mon Jun 05, 2017 11:00 pm
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.
Top
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Mon Jun 05, 2017 11:01 pm
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.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8085
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

Re: Probablity/P&C problem Exam Pack1

by Scott@TargetTestPrep » Mon Mar 02, 2020 5:55 am
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1