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If two integers between -5 and 3, inclusive, are chosen at random, whi

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

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New post 12 Jun 2018, 23:14
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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

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

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New post 12 Jun 2018, 23:30
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Bunuel wrote:
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


Questions involving number properties can often be solved logically, without using equations.
This is a Logical approach.

We have a total of 9 numbers, 5 odd and 4 even
A. occurs when both integers are odd or both are even this is 5C2 + 4C2 = 10 + 6 = 16
B. occurs when one is odd and the other even this is 5C1*4C1 = 5*4 = 20
C. occurs when at least one is even = 4C2 + 4C1*5C1 = 6 + 20 = 26
D. occurs when both are odd = 5C2 = 10
E. occurs when one is positive and the other negative. = 5C1*2C1 = 5*2 = 10


So (C) is our answer.
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If two integers between -5 and 3, inclusive, are chosen at random, whi  [#permalink]

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New post 12 Jun 2018, 23:31
The list of numbers are {-5, -4, -3, -2, -1, 0, 1, 2, 3}, since it is asked as most likely to be true. I calculated the possible combinations to arrive at an answer.

Even numbers of the set (-4, -2, 0, 2) = 4 numbers
Odd numbers of the set (-5, -3, -1, 1, 3) = 5 numbers
Negative numbers of the set (-5, -4, -3, -2, -1) = 5 numbers
Positive numbers (1, 2, 3) = 3 numbers

A. The sum of the two integers is even

For the sum to be even both number should be even or both should be odd.

4C2 + 5C2 = 6+10 = 16


B. The sum of the two integers is odd

The sum to be odd one should be even and one should be odd.
4C1 * 5C1 = 20


C. The product of the two integers is even

Product to be even both should be even or one should be even.

4C2 + (4C1*5C1) = 26


D. The product of the two integers is odd

product to be odd both should be odd.
5C2 = 10


E. The product of the two integers is negative

product to be negative. one should be negative and other should be positive.
5C1 * 3C1 = 15


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

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New post 16 Jun 2018, 10:12
1
Bunuel wrote:
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


The integers are:

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

We have 5 negative numbers, 1 zero, and 3 positive numbers.

We have 5 evens and 4 odds.

Since even x even = even and even x odd = even, it’s most likely to be true that we will get an even product.

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

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New post 27 Jun 2018, 10:27
DavidTutorexamPAL wrote:
Bunuel wrote:
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


Questions involving number properties can often be solved logically, without using equations.
This is a Logical approach.

We have a total of 9 numbers, 5 odd and 4 even
A. occurs when both integers are odd or both are even this is 5C2 + 4C2 = 10 + 6 = 16
B. occurs when one is odd and the other even this is 5C1*4C1 = 5*4 = 20
C. occurs when at least one is even = 4C2 + 4C1*5C1 = 6 + 20 = 26
D. occurs when both are odd = 5C2 = 10
E. occurs when one is positive and the other negative. = 5C1*2C1 = 5*2 = 10


So (C) is our answer.





Hi

for C, I calculated as below

occurs when at least one is even- 4C1 * 8C1 (select 1 even int and select 1 from remaining 8 integers ) =32

I too got answer C , but what's wrong with my method ?
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If 2 integers between -5 and 3, inclusive are chosen at random, which  [#permalink]

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New post 03 Sep 2018, 09:10
If 2 integers between -5 and 3, inclusive are chosen at random, which of the following is most likely to be true?

A. The sum of two integers is even
B. The sum of two integers is odd
C. The product of two integers is even
D. The product of two integers is odd
E. The product of two integers is even negative
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If two integers between -5 and 3, inclusive, are chosen at random, whi  [#permalink]

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New post Updated on: 03 Sep 2018, 22:50
ManasviHP wrote:
If 2 integers between -5 and 3, inclusive are chosen at random, which of the following is most likely to be true?

A. The sum of two integers is even
B. The sum of two integers is odd
C. The product of two integers is even
D. The product of two integers is odd
E. The product of two integers is even negative

Posted from my mobile device


OA: C

Total Number of integers between -5 and 3 = 9 (-5,-4,-3,-2,-1,0,1,2,3)
Number of Even Integers between -5 and 3 =4 (-4,-2,0,2)
Number of Odd Integers between -5 and 3 =5 (-5,-3,-1,1,3)

A. The sum of two integers is even
Sum of Two integers will be even in 2 conditions
Case 1: EVEN+EVEN
We have to find the number of ways of selecting 2 even integers.
\(C(4,2) = \frac{4!}{(4-2)!2!}=\frac{4!}{2!2!}=6\)
Case 2:ODD+ODD
We have to find the number of ways of selecting 2 Odd integers.
\(C(5,2) = \frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=10\)
Total favourable case\(= 10+6 = 16\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability\(= \frac{16}{36} =\frac{4}{9}=0.44\)

B. The sum of two integers is odd
Sum of Two integers will be even in 1 conditions
Case 1: EVEN+ODD
We have to find the number of ways of selecting 1 even integers and 1 odd integer.
\(C(4,1)*C(5,1)= \frac{4!}{(4-1)!1!}*\frac{5!}{(5-1)!1!}=\frac{4!}{3!1!}*\frac{5!}{4!1!}=20\)
Total favourable case\(=20\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability\(= \frac{20}{36} =\frac{5}{9}=0.55\)

C.The product of two integers is even
The probability of product of two integers being even = 1 - Probability of product of two integers being Odd
Product of two integers will be odd when both are odd
Case :ODD*ODD
We have to find the number of ways of selecting 2 Odd integers.
\(C(5,2) = \frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=10\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability of product of two integers being Odd \(= \frac{10}{36} = \frac{5}{18}\)
Probability of product of two integers being Even\(=1-\frac{5}{18} = \frac{13}{18}=0.72\)(Most likely to be true)

D. The product of two integers is odd
Probability of product of two integers being Odd \(= \frac{10}{36} = \frac{5}{18}=0.27\)

E.The product of two integers is even negative
Possible combinations such that product of two integers is even negative :9 {(-4,1);(-4,2);(-4,3);(-2,1);(-2,2);(-2,3);(2,-1);(2,-3);(2,-5)}
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability of product of two integers is even negative \(= \frac{9}{36} = \frac{1}{4}=0.25\)

Edit: E.The product of the two integers is negative
Case: -ve integer(5 choices) * +ve integer(3 choices)
We have find one negative integer and one positive integer
Number of Selecting one negative integer and one positive integer\(= C(5,1)*C(3,1) = 5*3=15\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability of product of the two integers is negative\(= \frac{15}{36}=\frac{5}{12}=0.42\)

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Originally posted by Princ on 03 Sep 2018, 09:59.
Last edited by Princ on 03 Sep 2018, 22:50, edited 1 time in total.
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Re: If two integers between -5 and 3, inclusive, are chosen at random, whi  [#permalink]

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New post 03 Sep 2018, 20:40
ManasviHP wrote:
If 2 integers between -5 and 3, inclusive are chosen at random, which of the following is most likely to be true?

A. The sum of two integers is even
B. The sum of two integers is odd
C. The product of two integers is even
D. The product of two integers is odd
E. The product of two integers is even negative


Merging topics. Please check the discussion above.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 03 Sep 2018, 22:55
If two integers between -5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?

So we have -5,-4,-3,-2,-1,0,1,2,3 choosing two integers = 9C2=36
Since the total possible cases in all choices is SAME 36, the choice which gives us the maximum cases in the given restriction is MOST likely
Choices A to D are dependent on Even and Odd, so let us see how many are even and how many odd..
even - 4 ; odd - 5


A) The sum of the two integers is even
Sum is even if both are even or both odd - 5C2+4C2=10+6=16

B) The sum of the two integers is odd
Sum is odd if one is even and other odd - 5C1*4C1=5*4=20

C) The product of the two integers is even
Product is even when both are even - 4C2 and when one is odd and other even - 5C1*4C1............Total = 4C2+5C1*4C1=6+5*4=26

D) The product of the two integers is odd
Product is odd if both odd - 5C2=10

E) The product of the two integers is negative
there are 5 negative and 3 positive, so cases = 5*3=15

so C gives us the max cases

Note - if you have got cases in A as 16, B will be 36-16=20 and
when you have got cases in C as 26, D will be 36-C=36-26=10

chetna14, "why C is not 4C1*8C1" is because there are repetitions when both 4C1 and 8C1 give us an even number
Example say 4C1 gives us 0 and 8C1 gives us 2 BUT there will be a case when 4C1 will give us 2 and 8C1 will give us 0, so set (0,2) is repeated similarly there are 4C2 cases so 32-4C2=32-6=26
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