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ps- VERY NICE- NUMBER PROPERTIES

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Manager
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ps- VERY NICE- NUMBER PROPERTIES [#permalink] New post 11 Jun 2007, 09:12
00:00
A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
If two integers between –5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?


(A) The sum of the two integers is even
(B) The sum of the two integers is odd
(C) The product of the two integers is even
(D) The product of the two integers is odd
(E) The product of the two integers is negative
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 [#permalink] New post 11 Jun 2007, 09:47
Lets analyze this one by one.

Between -5 and 3 (inclusive) there are 6 odd numbers -5.-3,-1,1,3 and there are 4 evens -4,-2,2

There are 5 negative and 3 positive numbers

I am tempted to intuitively say that product of 2 integers being even seems to be the best bet, but lets check the probabilities in detail :

A) happens when both numbers are odd or both are even : 6/10*5/9 + 4/10*3/9 = 7/15
B) happens if one odd and one even 6/10*4/9+4/10*6/9=9/15
C)happens when one of the numbers is even = 1- P(both are odd)
= 1 - ( 6/10 * 5/9) = 1-1/3 - 2/3 = 10/15
D)happens when both numbers are odd = 6/10 * 5/9 =1/3 or 5/15
E) happens when one of the numbers is <0>=0) = 1 - (4/10*3/9) = 13/15 --> AHA moment

Ans E , Is this the OA ?
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 [#permalink] New post 11 Jun 2007, 10:29
From a first glence, I'd pick C

Analyzing the problem:

Set: [-5, -4, -3, -2, -1, 0, 1, 2, 3] (total 9 integers)

# negatives: 5
# positives: 3
# Zeros: 1 :roll:
# Evens: 4
# Odds: 5

Now, lets calculate the probablity or the likelyhood of each event:

(A) The sum of the two integers is even
That'll be when picking 2 odds or 2 evens
P(A)= 5/9 x 4/8 + 4/9 x 3/8 = 32/72 = 4/9

(B) The sum of the two integers is odd
That holds when picking 1 odd and 1 even
P(B) = 5/9 x 4/8 + 4/9 x 5/8 = 2 x 20/72 = 40/72 = 5/9

(C) The product of the two integers is even
That'll be true when picking eithe 1 even 1 odd or 2 evens.
P(C) = P(B) + 4/9 x 3/8 = 5/9 + 12/72 = 5/9 + 1/6 = 39/54

(D) The product of the two integers is odd
That holds when picking two odds only ..
P(D) = 5/9 x 4/8 = 20/72 = 5/18 = 2.5/9

(E) The product of the two integers is negative
that holds when picking one negative and one positive
P(E) = 5/9 x 3/8 + 3/9 x 5/8 = 30/72 = 5/12


Yahoooo. My first glence worked out perfectly and my stats courses in college are helping me now :lol: ..
39/54 has the largest value among all answer choices.

Answer: C
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 [#permalink] New post 11 Jun 2007, 10:36
great problem. where did you get this from?
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 [#permalink] New post 11 Jun 2007, 10:38
[-5,-4,-3,-2,-1,0,1,2,3] = total numbers = 9 (zero is even !!)

[o,e,o,e,o,e,o,e,o]

The sum of the two integers is even

(4/9*3/8) + (5/9*4/8) = (choosing even, even) + (choosing odd, odd) = 32/72

The sum of the two integers is odd

(5/9*4/8) + (4/9*5/8) = (choosing odd,even) + (choosing even, odd) = 40/72

or simpler way 1-(32/72) = 40/72

The product of the two integers is even

(4/9*3/8) + (4/9*5/8) + (5/9*4/8) = (choosing even, even) + (choosing even, odd) +(choosing odd, even) = 52/72

The product of the two integers is odd

simpler way 1-(52/72) = 20/72 (since the outcome has to be odd or even)

The product of the two integers is negative

(5/9*4/8) + (4/9*5/8) = (choosing positive, negative) + (choosing negative, positive) = 40/72

the answer is 52/72 - answer (C)

:-D
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 [#permalink] New post 11 Jun 2007, 10:53
uhhhh ... I am kicking myself for not being able to count :(. I hope I dont do this on test day which is a week away now.

Goes to show you why doing problems at 3 am is not a good idea...

Great problem though.
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 [#permalink] New post 11 Jun 2007, 12:49
oa is C and I got the problem from a course which name I can't tell for copyright reasons....I also got E first time
  [#permalink] 11 Jun 2007, 12:49
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