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If the variables w,x,y and z are chosen at random so that
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Updated on: 20 Apr 2007, 09:14
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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
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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.
Re: If the variables w,x,y and z are chosen at random so that
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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.
Re: If the variables w,x,y and z are chosen at random so that
[#permalink]
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?
Re: If the variables w,x,y and z are chosen at random so that
[#permalink]
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.
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.
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!
gmatclubot
Re: If the variables w,x,y and z are chosen at random so that [#permalink]