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Probability Query

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Nijo Rising GMAT Star Default Avatar
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Probability Query Post Wed Aug 13, 2014 1:27 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If 2 of the expressions x-y, x+y, x+5y 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

    OA is E

    Thanks

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    Post Wed Aug 13, 2014 2:00 am
    Nijo wrote:
    If 2 of the expressions x-y, x+y, x+5y 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
    FOIL every possible pair:

    (x-y)(x+y) = x² - y²
    (x-y)(x+5y) = x² + 4xy - 5y²
    (x-y)(5x-y) = x² - 6xy + y²
    (x+y)(x+5y) = x² + 6xy + 5y²
    (x+y)(5x-y) = 5x² + 4xy - y²
    (x+5y)(5x-y) = 5x² + 24xy - 5y²

    Of the 6 results, only the first is in the required form:
    x² - y² = x² - (1y)², where b=1.

    Thus:
    P(the product will be in the required form) = 1/6.

    The correct answer is E.

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    Post Wed Aug 13, 2014 4:40 am
    Quote:
    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 x² - (by)², where b is an integer?

    A) 1/2
    B) 1/3
    C) 1/4
    D) 1/5
    E) 1/6
    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)

    -----------

    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, we have a free video on calculating combinations (like 4C2) in your head: http://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

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    Post Wed Aug 13, 2014 7:29 am
    Nijo wrote:
    If 2 of the expressions x-y, x+y, x+5y 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

    Thanks
    If we Choose to go by Options...

    There are 4 Expressions given x-y, x+y, x+5y and 5x-y

    The total ways of combination of two out of 4 expressions = 4C2 = 6

    i.e. Denominator of the probability can be either 6 or any factor of 6 eliminating options C and D

    Now (5x-y) when multiplied by any other expression given will not give us any result of the form x² - (by)²

    therefore we have to check the possible out comes of the product of expressions x+y, x+5y and x-y

    Out of 3 combinations that we can make using them, we can easily identify that only one product of two of these expressions x+y and x-y will give you the desired result.

    Therefore Favorable outcome = 1

    Probability = 1/6

    Answer: Option E

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    Post Wed Aug 13, 2014 7:43 pm
    Its simple. the coefficient of 'x' in the question is 1.

    so (x+5y), (5x -y) cannot be multiplied with any of the other options,

    so that leaves only one combination, (x+y) * (x-y). Since there are 4 items, there are 4C2 ways of choosing the outcomes. i.e 6 ways.

    so answer is 1/6.

    hope it makes sense.

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    Post Tue Aug 09, 2016 1:22 pm
    Brent@GMATPrepNow wrote:
    Quote:
    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 x² - (by)², where b is an integer?

    A) 1/2
    B) 1/3
    C) 1/4
    D) 1/5
    E) 1/6
    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)

    -----------

    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, we have a free video on calculating combinations (like 4C2) in your head: http://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
    Hi Brent, how about I solve it considering the probability of choosing algebraic expressions 1 and 2 ( x-y and x+y) out of the 4 expressions, so I would have that P=((1/4)+(1/4))*(1/3). Would it be correct?

    Post Tue Aug 09, 2016 7:45 pm
    Hi All,

    This question is based heavily on algebra patterns. If you can spot the patterns involved, then you can save some time; even if you can't spot it though, a bit of 'brute force' math will still get you the solution.

    We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

    Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes. From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

    By brute-forcing the 6 possibilities, you would have...
    (X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
    (X+Y)(X-Y) = X^2 - Y^2
    (X+Y)(5X-Y) = X^2 + 4XY - Y^2
    (X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
    (X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
    (X-Y)(5X-Y) = 5X^2 -6XY + Y^2

    Only the second option is in the proper format, so we have one option out of six total options.

    Final Answer: E

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    Post Fri Aug 19, 2016 1:42 am
    This one is asked pretty often, so don't feel bad for struggling with it!

    x² - (by)² factors as

    (x + by) * (x - by)

    So this will only work if both parts of the pair have the same coefficient on y. Given our four possibilities, the only pair that works is (x + y) * (x - y), since they each have 1y. So only one pair out of (4 choose 2), or 6 total, works, giving us 1/6.

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