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If the variables w,x,y and z are chosen at random so that [#permalink]
20 Apr 2007, 06:12

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C

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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

Last edited by kevincan on 20 Apr 2007, 08:14, edited 1 time in total.

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, 10:26, edited 1 time in total.

Re: PS: Product of 0 [#permalink]
20 Apr 2007, 18:15

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: PS: Product of 0 [#permalink]
20 Apr 2007, 23:51

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

gmatclubot

Re: PS: Product of 0
[#permalink]
20 Apr 2007, 23:51