Author 
Message 
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
Location: Madrid

If the variables w,x,y and z are chosen at random so that [#permalink]
Show Tags
20 Apr 2007, 07:12
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum.
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)(y1)(z+2) is not equal to 0?
(A) 1/4 (B) 7/24 (C) 1/3 (D) 3/8 (E) 5/12
Last edited by kevincan on 20 Apr 2007, 09:14, edited 1 time in total.



Intern
Joined: 08 May 2006
Posts: 24

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.



GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
Location: Madrid

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.



Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

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 ?
Last edited by Mishari on 20 Apr 2007, 11:26, edited 1 time in total.



Director
Joined: 14 Jan 2007
Posts: 774

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 = 62= 4(as 2 cases already counted in w = 0)
y =1 , number of cases = 62 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'



VP
Joined: 08 Jun 2005
Posts: 1145

hi
Total  4! = 24
Probability mass function
for k(from 1 to n) = C(n,k)*(nk)!*(1^k1)
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
2412+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:
http://en.wikipedia.org/wiki/Binomial_distribution



Manager
Joined: 28 Feb 2007
Posts: 197
Location: California

Re: PS: Product of 0 [#permalink]
Show Tags
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)(y1)(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?



GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262
Location: Madrid

Re: PS: Product of 0 [#permalink]
Show Tags
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)(y1)(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




Re: PS: Product of 0
[#permalink]
21 Apr 2007, 00:51






