If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of 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

Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);

\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.

Answer: E.
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Okay, first recognize that x² - (by)² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
x² - 25y²
4x² - 9y²
49m² - 100k²

In general, we can factor differences of squares as follows:
a² - b² = (a-b)(a+b)

So . . .
x² - 25y² = (x+5y)(x-5y)
4x² - 9y² = (2x+3y)(2x-3y)
49m² - 100k² = (7m+10k)(7m-10k)

----OKAY, ONTO THE QUESTION------------------

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)

If anyone is interested, I have added a video (below) on calculating combinations (like 4C2) in your head

# 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

Answer: E

RELATED VIDEO

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it should be 1/2 * 1/3 = 1/6

First case you can chose either x+y or x-y. So you have 2 options out of 4. second time you have 1 option out of 3

The only way to get x^2 - (by)^2 is if you multiply (x+by) with (x-by). Looking at the expressions given, the only pair that will satisfy this is (x+y) & (x-y)

No of ways to pick pairs = C(4,2) = 6

No of pairs satisfying condition = 1

Probability = 1/6
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Hey guys,

I know its a stupid question, but for some reason i cannot figure it out now... Maybe Im just too tired...

Why cant we do this questions simpler? as in -

1/4*1/3?

I understand perfectly the 4C2 way, but i want to understand the others as well...

Thanks.

Sry for the stupidity...

How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8.

Question says: "If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random..." so, at least for me, it's quite obvious that there is no replacement whatsoever. Also if it were the case the question would explicitly mention that.
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How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

Don't these two problems have the same concept?

Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.
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Hello Bunuel,

Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused.

Since the wanted product is (x + y ) (x - y), I assumed that I will be picking (x+y) then (x-y) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did.

Thank you very much Bunuel.

It should be 1/4*1/3*2, since you can choose (x+y) first and then (x-y) or vise versa.
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This is how I analysed this. x-y is a must here since the form asked is x^2-by^2. So probabability of selecting x-y is 1/4 and probability of selecting the rest is 2/3. the probability of both is 1/4*2/3. Is this right???

I was hoping someone can help me understand this. What is the 4C2 way? C means what?

How I thought of the problem was each equation has a 1/4 chance

1/4 1/4 1/4 1/4 - Two out of the 4 equations = success so the probability of getting one of the equations we need is 2/4 or 1/2
then we have 3 equations left over, and the probability of getting the second equation we need is 1/3
so the probability of getting both equations is 1/2*1/3 = 1/6

However I only got this after you guys said that we needed a x+y, x-y

The b in the question threw me off, I am not sure that if I was given a similar problem some weeks from now I would understand what to do with that b

Akuma,

4C2 stands for binomial coefficient with parameters 4 and 2. In layman's terms, it says how many unique combinations of two can I derive from a set of 4 items. In mathematical terms, it simply means 4! / (2! x (4-2)!) which simplifies to 4! / (2! x 2!) which is equivalent to 6. Now you just need to figure out how many of the combinations fulfill your criteria - in this case 1 - and place that number over 6 to get the probability. To get the amount of unique pairings of Y numbers in a set of X (where order does not matter) I believe you can always use this shorthand formula

Total No. of ways of selecting 2 items from 4 items = 4C2 = 6. We can see from the 4 items given that only one pair when multiplied will be of the form (x+ay)(x-ay). So answer is 1/6. E.

Kudos Please... If my post helped.

VERY SIMPLE WAY TO SOLVE THIS PROB WITHOUT ANY COMPLEX THEORY N ANY CALCULATION :
FROM THE QUESTION STEM , YOU COULD UNDERSTAND VERY EASILY THAT ONLY ONE COMBINATION I.E (X+Y) AND (X-Y) IS THERE TO GET X2-(BY)2
SO WE HAVE TO CHOOSE ONLY (X+Y) AND (X-Y)
wE CAN SELECT IT IN TWO WAY I.E . (X+Y) AND THEN (x-Y) OR (X-Y) AND THEN (X+Y)
FOR FIRST CASE PROBABILITY =(1/4*1/3)
FOR SECOND CASE PROBABILITY =(1/4*1/3)
sO TOTAL PROBABILITY =2*((1/4*1/3)=1/6
HOPE IT COULD SIMPLY WORK FOR YOU

how many ways to select a pair from 4 expressions?
\(\frac{4!}{2!2!}=6\)

how many pairs to achieve \(x^2-(by)^2\)?
(x+y)(x-y) only or 1 pair

\(\frac{desired}{{all pair possibilities}}=\frac{1}{6}\)

I made a careless error on this problem...

I thought that b(y^2) would be the same as (by)^2.... but nope... i didn't catch the parentheses.

Great explanations in this topic. Again, when explained, easier than it appeared.

As there are 2 steps to this (number of possible outcomes, number of valid pairs); if you're lost at figuring out the latter it is maybe of some use that it is easy to realize that if there are 6 possible outcomes, only 1/2, 1/3 and 1/6 could be valid answers as x/6 can't be simplified to 1/4 or 1/5. Now you only need to figure out if there are 1, 2 or 3 valid pairs, or if time ran out guess 1 out of 3.