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

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If the variables w,x,y and z are chosen at random so that [#permalink] New post 20 Apr 2007, 06:12
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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.
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 [#permalink] New post 20 Apr 2007, 07:42
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.
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 [#permalink] New post 20 Apr 2007, 08:15
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.
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 [#permalink] New post 20 Apr 2007, 08:19
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.
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 [#permalink] New post 20 Apr 2007, 09:03
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'
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re: [#permalink] New post 20 Apr 2007, 09:06
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:

http://en.wikipedia.org/wiki/Binomial_distribution
:)
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Re: PS: Product of 0 [#permalink] New post 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?
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Re: PS: Product of 0 [#permalink] New post 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.

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] 20 Apr 2007, 23:51
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