soni_pallavi wrote:Q1) If two of the four expressions x+y, x+5y ,x-y and 5x-y are chosen at random,what is the probability that product will be in the form of x^2 - (by)^2 ,where b is an integer?

a)1/2
b)1/3
c)1/4
d)1/5
e)1/6

Ans E

soni, please use the spoiler function to hide the correct answer so that others can attempt the question.

Okay, first recognize that x^2 - (by)^2 is a difference of squares.
Here are some examples of differences of squares:
x^2 - 25y^2
4x^2 - 9y^2
49m^2 - 100k^2
etc.

In general, we can factor differences of squares as follows:
a^2 - b^2 = (a-b)(a+b)

So . . .
x^2 - 25y^2 = (x+5y)(x-5y)
4x^2 - 9y^2 = (2x+3y)(2x-3y)
49m^2 - 100k^2 = (7m+10k)(7m-10k)

So, from the 4 expressions (x+y, x+5y ,x-y and 5x-y), only one pair (x+y and x-y) will result in a difference of squares when multiplied.

So, the question now becomes:
If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(both selected) = [# of outcomes in which x+y and x-y are both selected]/[total # of outcomes]

As always, we'll begin with the denominator.

total # of outcomes
There are 4 expressions, and we must select 2 of them.
Since the order of the selected expressions does not matter, we can use combinations to answer this.
We can select 2 expressions from 4 expressions in 4C2 ways (= 6 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

# of outcomes in which x+y and x-y are both selected
There is only 1 way to select both x+y and x-y

So, P(both selected) = 1/6 = E

Cheers,
Brent

We can also solve this question using probability rules. We'll begin at:

If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(x+y and x-y are both selected) = P(one of the required expressions is chosen 1st AND the other expression is chosen 2nd)
= P(one of the required expressions is chosen 1st) x P(the other expression is chosen 2nd)
= (2/4)(1/3)
= 1/6 = E

Cheers,
Brent

subhakam wrote:Hello Brent - don't we also calculate probability of the other expression chosen first and 1st expression chosen second?
P(x+y and x-y are both selected) = P(one of the required expressions is chosen 1st AND the other expression is chosen 2nd)
= P(one of the required expressions is chosen 1st) x P(the other expression is chosen 2nd)
= (2/4)(1/3)
= 1/6
OR
P(x-y and x+y) = P(x-y chosen first) and P(X+Y chosen second)= 1/6
Therefore final outcome is 1/6+ 1/6 = 1/3?? I am confused as to why we are not calculating the probability of P(X+Y and x-y) OR P (x-y and x+y)

Can you please explain ?

Many thanks

Regarding your question in green. Great question. By setting up my probability as P(one of the required expressions is chosen 1st AND the other expression is chosen 2nd), I've actually taken care of both possible scenarios.
Writing, "one of the required expressions" includes x+y chosen first, and it includes x-y chosen first.

The great thing about probability questions is that we can set up the solution a different way and still get the correct answer. Here's another approach:

P(x+y and x-y are both selected) = P(x+y chosen 1st AND x-y chosen 2nd OR x-y chosen 1st AND x+y chosen 2nd)
= P(x+y chosen 1st AND x-y chosen 2nd) + P(x-y chosen 1st AND x+y chosen 2nd)
= [P(x+y chosen 1st) X P(x-y chosen 2nd)] + [P(x-y chosen 1st) X P(x+y chosen 2nd)]
= [(1/4) X (1/3)] + [(1/4) X (1/3)]
= 1/12 + 1/12
= 1/6 = E

Cheers,
Brent