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Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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If two integers between –5 and 3, inclusive, are chosen at random, whi  [#permalink]

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4 00:00

Difficulty:   95% (hard)

Question Stats: 44% (02:28) correct 56% (02:29) wrong based on 120 sessions

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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.

OA is C

Pls verify reasoning.

The numbers are {-5,-4,-3,-2,-1,0,1,2,3}
Total = 9 and 5 odds 4 evens.

We don't have to divide by 9C2 to compare. Just compare the number of favorable outcomes.

1. Both are odd or both are even. Ways to get both odd or both even = 5C2 + 4C2 = 10 + 6 = 16

2. Sum is odd when - one is odd and one is even = 5C1 * 4C1 = 20

3. Either both are even or one is even and one odd = 4C2 + 5C1 * 4C1 = 6 + 20 = 26

4. Both are odd integers - 5C2 = 10

5. One is positive integer and one is negative integer = 5C1 * 3C1 = 15
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Re: If two integers between –5 and 3, inclusive, are chosen at random, whi  [#permalink]

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1. The sum of the two integers is even and 2 sum of two integers odd

for sum to be odd -- 5c1*4c1/9c2 = 20/36 thus for even 1-(20/36) = 16/36

3,4 product of two integers even/odd --

for integers to be odd ---- 5c2/9c2 = 52/72 for even 1-52/72 = 20/72

5 for product to be negative -- 0 will not be taken into account. Hence 8 numbers is the set.

5c1 * 3c1/ 8c2 = 15/28

thus comparison is between b 20/36, c 52/72 e 15/28

C is greatest since 52 is much closer to 72 hence much greater than 1/2. 20/36 and 15/28 are much closer to 1/2 though.

Thus C.

Good question.
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Re: If two integers between –5 and 3, inclusive, are chosen at random, whi  [#permalink]

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1
-5,-4,-3,-2,-1,0,1,2,3

Total cases = 9C2

1) EE or OO -> 4C1 * 3C1 + 5C1 * 4C1 = 12 + 20 = 32

2) EO -> 4C1 * 5C1 = 20

3) E*O + O*E + E*E (this is bigger than 1)

4) OO -> (Least of all)

5) -ve * + ve + -ve * +ve = 5C1 * 3C1 + 3C1 * 5C1 = 15 + 15 = 30

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Re: If two integers between –5 and 3, inclusive, are chosen at random, whi  [#permalink]

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3
gmat1220 wrote:
If two integers between –5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?

The sum of the two integers is even
The sum of the two integers is odd.
The product of the two integers is even.
The product of the two integers is odd.
The product of the two integers is negative.

Hi, MOST likely means we are looking for the MAX probability in all choices..
although it can be seen that (3) should be the answer.. Lets solve it
So before we start lets see what do these integers consist of and would help in eliminating choices..
-5 to 3 = -5, -4, -3, -2, -1, 0, 1, 2, 3..
Even = 4, ODD =5..
-ive = 5, +ive = 3 and one 0..

choosing two numbers will be same for all, so we can compare the ways for each..

lets see the choices now--

(1) The sum of the two integers is even
even +even is even and odd + odd is EVEN..
Prob = choose two even or two odd = 4C2 + 5C2 = 6 +10 =16

(2) The sum of the two integers is odd.
even +odd = odd..
so 5C1*4C1 = 5*4 = 20..

(3)The product of the two integers is even.
one number even and other any of remainig..
so 4C1 * 8C1 = 32

(4) The product of the two integers is odd.

both have to be odd = 5C2= 10

(5) The product of the two integers is negative.
one negative and one positive = 5*3 =15.

so (3) has the max ways
ans C
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Re: If two integers between –5 and 3, inclusive, are chosen at random, whi  [#permalink]

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_________________ Re: If two integers between –5 and 3, inclusive, are chosen at random, whi   [#permalink] 03 Sep 2018, 02:17
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