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# If two integers between -5 and 3, inclusive, are chosen at r

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If two integers between -5 and 3, inclusive, are chosen at r [#permalink]  12 Jun 2018, 07:33
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12% (01:54) correct 87% (01:29) wrong based on 16 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
[Reveal] Spoiler: OA

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Re: If two integers between -5 and 3, inclusive, are chosen at r [#permalink]  13 Jun 2018, 22:59
Need clear explaination plz.. Carcass
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Re: If two integers between -5 and 3, inclusive, are chosen at r [#permalink]  15 Jun 2018, 00:32
2
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Start with making the set of numbers $${-5,-4,-3,-2,-1,0,1,2,3}$$

The question stem states that which of the following is likely to be true. Hence it is possible that all of the options can occur however we are required to chose the option that has the highest probability.

Option A.
When the sum of two integers is even it means that either both are even or both are odd.

In the list
even numbers = -4,-2,0,2 = 4
odd numbers = -5,-3,-1,1,3 = 5

Hence for even number a possibility is when even is added with even, that would be $$\frac{4}{9}$$ * $$\frac{4}{9}$$ = $$\frac{16}{81}$$ (From 9 choices 4 are possible choices)
or for even number a possibility is when odd is added with odd $$\frac{5}{9}$$ * $$\frac{5}{9}$$ = $$\frac{25}{81}$$
Total probability = $$\frac{41}{81}$$ to get even number.

Let us tackle option B
Since sum of two integers from the same set can be either odd or even and we already have the probability for even,
the sum resulting in odd = 1 - $$\frac{41}{81}$$= $$\frac{40}{81}$$

we know at this point that B is not the correct ans

Let us tackle option C
There is a similar theme of odd and even for option c and d hence instead of finding the probability for even we should find the probability for odd. Since there is limited possibility for odd product i.e only when odd is multiplied with odd.
For odd the choice would be odd for the first pick and odd for the second pick hence, $$\frac{5}{9}$$ * $$\frac{5}{9}$$ = $$\frac{25}{81}$$. This probability is lower than choice A so option D is out
For option c the probability is 1 - $$\frac{25}{81}$$ = 56/81. This is more than option A hence so far option C is the candidate for right ans.

Finally option E.
This is slightly different because we need to work with +ve and -ve.

There are 5 negative numbers and 3 +ve numbers. 0 is neutral.
Hence for getting a negative number when two integers are multiplied it has to be either negative * positive or positive * negative.
negative * positive = $$\frac{5}{9} * \frac{3}{9} = \frac{15}{81}$$
positive * negative = $$\frac{3}{9} * \frac{5}{9}$$ = $$\frac{15}{81}$$
Therefore the probability is $$\frac{30}{81}$$ which is less than the probability for option c
Option C is the correct ans.
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: If two integers between -5 and 3, inclusive, are chosen at r   [#permalink] 15 Jun 2018, 00:32
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