NEW HERE? LEARN MORE
closeclose
Last visit was: 27 Apr 2024, 21:45 It is currently 27 Apr 2024, 21:45
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If the variables w,x,y and z are chosen at random so that

User avatar
B
kevincan
GMAT Instructor
Joined: 04 Jul 2006
Posts: 960
Own Kudos [?]: 693 [0]
Given Kudos: 6
Location: Madrid
 Q51  V50
Send PM
If the variables w,x,y and z are chosen at random so that [#permalink] New post   Updated on: 20 Apr 2007, 09:14
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hide Show timer Statistics
If the variables w,x,y and z are chosen at random so that each variable is a distinct element of the set {-2,-1,0,1}, what is the probability that the product w(x+1)(y-1)(z+2) is not equal to 0?

(A) 1/4 (B) 7/24 (C) 1/3 (D) 3/8 (E) 5/12



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!

Originally posted by kevincan on 20 Apr 2007, 07:12.
Last edited by kevincan on 20 Apr 2007, 09:14, edited 1 time in total.
avatar
doc14
Intern
Intern
Joined: 08 May 2006
Posts: 15
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   20 Apr 2007, 08:42
Does "distinct" mean that w,x,y,z must be different?
If so I still get a different answer to those listed: 15/24

24 possible outcomes of which 15 have either w =0, x=-1,y =1 or z =2.

Can anyone else get the right answer and show technique?

Thanks.
User avatar
B
kevincan
GMAT Instructor
Joined: 04 Jul 2006
Posts: 960
Own Kudos [?]: 693 [0]
Given Kudos: 6
Location: Madrid
 Q51  V50
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   20 Apr 2007, 09:15
doc14 wrote:
Does "distinct" mean that w,x,y,z must be different?
If so I still get a different answer to those listed: 15/24

24 possible outcomes of which 15 have either w =0, x=-1,y =1 or z =2.

Can anyone else get the right answer and show technique?

Thanks.


You're right! The question has been corrected.
User avatar
Mishari
Senior Manager
Senior Manager
Joined: 30 Nov 2006
Posts: 352
Own Kudos [?]: 891 [0]
Given Kudos: 0
Location: Kuwait
Concentration: Strategy - Finance
 Q49  V36
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   Updated on: 20 Apr 2007, 11:26
How many ways can we arrange the elements of the set over the four variabes = 4! = 4x3x2x1 = 24
The cases when the equation would equal zero is:
w = 0 --> 6 ways
x = -1 --> 5 ways
y = +1 --> 3 ways
z = -2 --> 2 ways
Sum: 16 ways

Probability that the equation equals zero = 16/24 = 2/3
Probability that the equation does NOT equal zero = 1 - 2/3 = 1/3

My Answer: C

What is OA ?

Originally posted by Mishari on 20 Apr 2007, 09:19.
Last edited by Mishari on 20 Apr 2007, 11:26, edited 1 time in total.
User avatar
vshaunak@gmail.com
Senior Manager
Senior Manager
Joined: 14 Jan 2007
Posts: 314
Own Kudos [?]: 902 [0]
Given Kudos: 0
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   20 Apr 2007, 10:03
Total combinations = 4! = 24

For the product to be zero, w = 0 or x = -1, y = 1 , z = -2

w = 0 , number of cases = 6

x = -1, number of cases = 6-2= 4(as 2 cases already counted in w = 0)

y =1 , number of cases = 6-2 -1 = 3 (as 2 cases already counted in w = 0 and 1 case already counted in x=-1)

z = -2 , number of cases = 1

Number of cases when product will be zero = 14
Probability of product to be zero = 14/24 = 7/12

Probability of product not to be zero = 5/12
Answer is 'E'
User avatar
KillerSquirrel
Director
Director
Joined: 08 Jun 2005
Posts: 523
Own Kudos [?]: 559 [0]
Given Kudos: 0
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   20 Apr 2007, 10:06
hi

Total - 4! = 24

Probability mass function

for k(from 1 to n) = C(n,k)*(n-k)!*(-1^k-1)

k=1 C(4,1)*3!*1 = 4*6*1 = 24

k=2 C(4,2)*2!*-1 = 6*2*-1 = -12

k=3 C(4,3)*1!*2 = 4*1*1 = 4

k=4 = 0

24-12+4 = 16

the Probability that equation equal 0 is: 16/24 = 2/3

the Probability that equation not equal 0 is: 1 - 2/3 = 1/3

answer is (C)

see also here for reference:

https://en.wikipedia.org/wiki/Binomial_distribution
:)
User avatar
techjanson
Manager
Manager
Joined: 28 Feb 2007
Posts: 104
Own Kudos [?]: 43 [0]
Given Kudos: 0
Location: California
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   20 Apr 2007, 19:15
kevincan wrote:
If the variables w,x,y and z are chosen at random so that each variable is a distinct element of the set {-2,-1,0,1}, what is the probability that the product w(x+1)(y-1)(z+2) is not equal to 0?

(A) 1/4 (B) 7/24 (C) 1/3 (D) 3/8 (E) 5/12


I got the C as answer. but my method was clumsy. anyone got any insights?
User avatar
B
kevincan
GMAT Instructor
Joined: 04 Jul 2006
Posts: 960
Own Kudos [?]: 693 [0]
Given Kudos: 6
Location: Madrid
 Q51  V50
Send PM
Re: If the variables w,x,y and z are chosen at random so that [#permalink] New post   21 Apr 2007, 00:51
kevincan wrote:
If the variables w,x,y and z are chosen at random so that each variable is a distinct element of the set {-2,-1,0,1}, what is the probability that the product w(x+1)(y-1)(z+2) is not equal to 0?

(A) 1/4 (B) 7/24 (C) 1/3 (D) 3/8 (E) 5/12


(w,x,y,z) must be a permutation of {-2,-1,0,1}. There are 4!=24 equally probable permutations.

For the product not to equal 0, w must not be 0, x must not be -1, y must not be 1 and z must not be -2.

w=-1, x=0, y=-2,z=1
w=-1, x=1, y=-2, x=0
w=-1, x=-2, y=0, x=1

w=1, x=0, y=-2, z=-1
w=1, x=-2, y=0, z=-1
w=1, x=-2, y=-1, z=0


w=-2,x=0,y=1,z=-1
w=-2,x=1,y=0,z=-1
w=-2,x=-2,y=0,z=1


So probability is 9/24= 3/8


We can think this way, w can be any of three numbers. For each possibility of w, consider the variable whose value has been taken by w. It can assume any of the 3 values left



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: If the variables w,x,y and z are chosen at random so that [#permalink]
21 Apr 2007, 00:51
Moderators:
Bunuel
Math Expert
92959 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions