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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
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Carcass wrote:
A number, x, is randomly selected from the integers from 42 to 92 inclusive.

Quantity A
Quantity B
The probability that x is odd.
The probability that x is even.


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given


Quick and dirty way of thinking about it. First off, we are pulling from the same set so denominator doesn't matter. The question becomes "are there more odd numbers or even numbers" well the interval both begins and ends with an even number. So there must be one more even than odd. So B is greater.
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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
1
You can solve this problem in less than five seconds.
The series begins and ends in even numbers and therefore there are more even numbers in the series.
So, the answer is B.

I explain the reasoning below.

Here's some shortcuts for comparing odds and evens (correct me if I'm wrong), given we're dealing with an arithmetic series

1) if the series starts and ends with the same type of number, there are more of that type
. in other words, if one type of number "surrounds" the other type, there are more of the surrounding type
. ex. if the series starts and ends with evens, if the evens "surround" the odds, there are more evens
.. ex. 2,3,4
.. this series starts and ends with evens, and therefore has more evens (2 evens, 1 odd)
. ex. if the series starts and ends with odds, there are more odds
.. ex. 1,2,3
.. this series starts and ends with odds, the odds "surround" the evens, and therefore the series has more odds (2 odds, 1 even)

2) If the series starts and ends with numbers of different types, then there are an equal number of odds and evens
. of if you prefer to think of it this way, neither type "surrounds" the other type
. ex. 1,2,3,4
.. series starts with one type (odd) and ends with another type (even), therefore series has equal number of each type (2 odds, 2 evens)
. ex. 2,3,4,5
.. series starts with one type (even) and ends with another type (odd), therefore series has equal number of each type (2 odds, 2 evens)

Extra:
If you know the total number of numbers, and the number of odds or evens, you can find the remaining # of the other type by subtracting
. ex. if we know we have 51 numbers in a series and 26 are even, then we know that 25 are odd (odd #s = total #s - even #s)
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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:
A number, x, is randomly selected from the integers from 42 to 92 inclusive.

Quantity A
Quantity B
The probability that x is odd.
The probability that x is even.


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given


----ASIDE---------------------
Since even and odd integers alternate, one might incorrectly assume that each probability = 1/2.
However, if there's an odd number of integers in the group (like 4, 5, 6, 7, 8), the number of even integers is one more than the number of odd integers.
----------------------------------

A nice rule says: the number of integers from x to y inclusive equals y - x + 1
So, the number of integers from 42 to 92 inclusive = 92 - 42 + 1 = 51

Since our integers start and end with an EVEN integer (42 and 92) there must be 26 EVEN integers and 25 ODD integers.

So P(x is EVEN) = 26/51
and P(x is ODD) = 25/51

We get:
QUANTITY A: 25/51
QUANTITY B: 26/51

Answer: B

Cheers,
Brent



Hey Brent, is this rule applicable for both odd and even integers? "However, if there's an odd number of integers in the group (like 4, 5, 6, 7, 8), the number of even integers is one more than the number of odd integers."

For instance, if we wanted to find whether the number of odd or even numbers are greater between 1 and 91, there would be more odd numbers right? (91-1+1 = 91)
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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
niveda94 wrote:
Hey Brent, is this rule applicable for both odd and even integers? "However, if there's an odd number of integers in the group (like 4, 5, 6, 7, 8), the number of even integers is one more than the number of odd integers."

For instance, if we wanted to find whether the number of odd or even numbers are greater between 1 and 91, there would be more odd numbers right? (91-1+1 = 91)


The formula: the number of integers from x to y inclusive equals y - x + 1
This applies only two consecutive integers

In order to for it to apply to consecutive ODD or EVEN integers, we'll need to make a small adjustment.

Let's say we want to find the number of ODD integers from 11 to 32 inclusive.
The above formula tells us that the number of integers = 38 - 11 + 1 = 28
So there are 28 integers in total.
Since the values begin with an ODD integer (11) and end with an EVEN integer (38), we know that half of the values are ODD and half are EVEN.
28/2 = 14
So there are 14 ODD integers from 11 to 32 inclusive, and there are 14 EVEN integers from 11 to 32 inclusive.


Let's try one more.
Let's say we want to find the number of EVEN integers from 66 to 96 inclusive.
The above formula tells us that the number of integers = 96 - 66 + 1 = 31
So there are 31 integers in total.
In this case, the values begin with an EVEN integer (66) and end with an EVEN integer (96).
This means the number of EVEN integers is 1 greater then the number of ODD integers.
So there are 15 ODD integers from 66 to 96 inclusive, and there are 16 EVEN integers from 66 to 96 inclusive.
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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
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Re: A number, x, is randomly selected from the integers from 42 [#permalink]
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