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Mathematical approach for Probability Question

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Mathematical approach for Probability Question

by newgmattest » Thu Jun 02, 2011 8:17 pm
Question as reference:

If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a2 - b2, where a and b are both positive integers?

A: 2/3
B: 1/2
C: 1/3
D: 1/4
E: 1/6

Now, the answer is A: 2/3.

Question is because here we just had 4 elements in set, we can list all possible options quickly. What is mathematical logic behind this to solve any problem with say 20 set elements?
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by Frankenstein » Thu Jun 02, 2011 8:39 pm
Hi,
1)Consider (n+1)^2-n^2 = 2n+1. So, any odd number(except 1) can be written as difference between squares of two positive integers in at least one way.So, if we choose any two odd numbers, their product is odd and can be written in the form a^2-b^2.
2)Consider (n+2)^-n^2 = 4n+4 =4(n+1). So, any multiple of 4(except 4) can be written as difference between squares of two positive integers.
So, if we choose two distinct even numbers the product is a multiple of 4 and greater than or equal to 8. So, their product can always be written in the form a^2-b^2
3)So, if we choose an odd number and even, their product is even. Now, if their product is multiple of 4 and greater than or equal to 8, it can be written in the form a^2-b^2.

Lets consider the same example {2,5,7,8}
1)2 odd number picked in 1 way
2)2 evens picked in 1 way
3)1 odd and 1 even picked such that their product is multiple of 4 in 2 ways
So total 4 ways out of 6 ways. Hence, probability is 4/6 = 2/3

I hope my post makes sense.
Last edited by Frankenstein on Thu Jun 02, 2011 8:41 pm, edited 1 time in total.
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by cans » Thu Jun 02, 2011 8:40 pm
a^2-b^2 = (a-b)(a+b)
Thus if we can break the product in the form p*q where either both p &q are even or both p&q are odd, we can write product in form of a^2-b^2
Thus you need to just check that.
product of m,n where m and n are even, will be considered always. product of 2 odd numbers will again be considered.
product of odd and even needs to be checked. (like 2*7 can't be written where as 5*8 can be written)
also if in product of odd and even, if even is multiple of 4, then again product can be written in form of (a^2-b^2)
Thus from total subtract the cases when 2 will be multiple by odd number.
like in this case, total=4C2=6
Cases when 2 is multiplied by odd number = 2*5,2*7 means 2
Thus valid = 6-2=4
prob = 4/6=2/3
:)
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by GMATGuruNY » Tue Jun 14, 2011 9:37 am
newgmattest wrote:Question as reference:

If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a2 - b2, where a and b are both positive integers?

A: 2/3
B: 1/2
C: 1/3
D: 1/4
E: 1/6

Now, the answer is A: 2/3.

Question is because here we just had 4 elements in set, we can list all possible options quickly. What is mathematical logic behind this to solve any problem with say 20 set elements?
I received a PM asking me to comment.

The solutions offered by Frankenstein and Cans are both great.

Here's one more way to look at the problem:

a² - b² = (a+b)(a-b).
The difference between the factors = (a+b)-(a-b) = 2b.
Since b is an integer, 2b is an even integer, implying that the difference between the factors is even.
For the difference to be even, both factors must be even or both must be odd (since even-even=even and odd-odd=even).
Thus, any integer that can be written as the product of two distinct even factors or two distinct odd factors can be represented in the form a² - b².

Any multiple of 4 greater than 4 can be written as the product of two distinct even factors.
Two ways to yield a multiple of 4:
Even*even.
Multiple of 4 * odd.

Thus, given a set of distinct positive integers, the following pairs will yield a product that can be written as a² - b²:
Every pair of even integers.
Every pair that includes a multiple of 4 and an odd integer.
Every pair of odd integers.

Thus, the following pairs will yield a product that cannot be written as a² - b²:
Every pair that includes an odd integer and an even integer that is not a multiple of 4.

The quickest approach might be to count the number of bad pairs.

Given (2,5,7,8):
Total possible pairs = 4C2 = 6.
2 bad pairs: (2,5) and (2,7).
Thus, number of good pairs = total-bad = 6-2 = 4.
P(good pair) = 4/6 = 2/3.

The correct answer is A.
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