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

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.

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

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

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 \(= \frac{9}{36} = \frac{1}{4}=0.25\)

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

We need to choose 2 numbers from {-5, -4, -3, -2, -1, 0, 1, 2, 3}

The correct answer choice will be the one that has the most possible outcomes

A. The sum of the two integers is even
There are 2 ways to get an EVEN sum:
1) Both integers are even
OR
2) Both integers are odd

1) We can choose 2 even integers in 4C2 ways = 6 ways (since there are 4 even integers to choose from)
2) We can choose 2 odd integers in 5C2 ways = 10 ways (since there are 5 odd integers to choose from)

TOTAL possible outcomes = 6 + 10 = 16


B. The sum of the two integers is odd
One number must be even, AND the other must be odd

We can choose 1 odd integer in 5 ways AND we can choose 1 even integer in 4 ways

TOTAL possible outcomes = 5 x 4 = 20


C. The product of the two integers is even
There are 2 ways to get an EVEN product:
1) Both integers are even
OR
2) One integer is even AND the other is odd

1) We can choose 2 even integers in 4C2 ways = 6 ways (since there are 4 even integers to choose from)
2) We can choose 1 even integer in 4 ways AND we can choose 1 odd integer in 5 ways. 4 x 5 = 20

TOTAL possible outcomes = 6 + 20 = 26


D. The product of the two integers is odd
There is one way to get an ODD product:
1) Both integers are odd

1) We can choose 2 odd integers in 5C2 ways = 10 ways

TOTAL possible outcomes = 10


E. The product of the two integers is negative
There is 1 way to get a NEGATIVE product:
1) One integer is negative AND the other is positive
We can choose 1 negative integer in 3 ways AND we can choose 1 positive integer in 5 ways.
TOTAL possible outcomes = 3 x 5 = 15

Answer: C

Cheers,
Brent

Bunuel DavidTutorexamPAL

Hi,

I have a conceptual query.

Let's take option a) for example.

It can only occur when either both integers are odd or both integers are even

Why isn't the total number of possibilities 4C1 X 4C1 + 5C1 X 5C1?

Why is it 4C2 + 5C2?

Hey smlprkh,

The difference is that the question assumes that the integers are different.
So your calculation also includes (1,1), for example, while 5C2 does not. Additionally, you do not take into account repetitions. For example, you count (1,3) and (3,1) as separate instances instead of the same one.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.