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There is a total of 120 marbles in a box, each of which is r

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There is a total of 120 marbles in a box, each of which is r Post Wed Aug 06, 2014 3:10 am
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    There is a total of 120 marbles in a box, each of which is red, green, blue, or white. If one marble is drawn from the box at random, the probability that it will be white is 1/4 and the probability that it will be green is 1/3. What is the probability that the marble will be either red or blue?

    (A) 1/6
    (B) 1/4
    (C) 2/7
    (D) 1/3
    (E) 5/12

    OA E

    Please explain

    Regards
    Sachin

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    Post Wed Aug 06, 2014 7:37 am
    sachin_yadav wrote:
    There is a total of 120 marbles in a box, each of which is red, green, blue, or white. If one marble is drawn from the box at random, the probability that it will be white is 1/4 and the probability that it will be green is 1/3. What is the probability that the marble will be either red or blue?

    (A) 1/6
    (B) 1/4
    (C) 2/7
    (D) 1/3
    (E) 5/12
    ...the probability that it [the marble] will be white is 1/4
    So, 1/4 of the 120 marbles are white.
    In other words, there are 30 white marbles.

    ...the probability that it [the marble] will be green is 1/3
    So, 1/3 of the 120 marbles are green .
    In other words, there are 40 green marbles.

    Altogether, there are 70 marbles that are EITHER white or green.
    This means that the remaining 50 marbles are EITHER red or blue.

    So, P(selected marble is EITHER red or blue) = 50/120 = 5/12

    Answer: E

    Cheers,
    Brent

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    Post Wed Aug 06, 2014 7:42 am
    sachin_yadav wrote:
    There is a total of 120 marbles in a box, each of which is red, green, blue, or white. If one marble is drawn from the box at random, the probability that it will be white is 1/4 and the probability that it will be green is 1/3. What is the probability that the marble will be either red or blue?

    (A) 1/6
    (B) 1/4
    (C) 2/7
    (D) 1/3
    (E) 5/12

    Alternatively, we can use the COMPLEMENT.
    That is, P(Event A happening) = 1 - P(Event A not happening)

    So, P(selected marble is EITHER red or blue) = 1 - P(selected marble is NEITHER red nor blue)
    = 1 - P(selected marble is either white or green)
    = 1 - [P(marble is white) + P(marble is green)]
    = 1 - [1/4 + 1/3]
    = 1 - [3/12 + 4/12]
    = 1 - 7/12
    = 5/12

    Answer: E

    Cheers,
    Brent

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    Post Wed Aug 06, 2014 12:02 pm
    sachin_yadav wrote:
    There is a total of 120 marbles in a box, each of which is red, green, blue, or white. If one marble is drawn from the box at random, the probability that it will be white is 1/4 and the probability that it will be green is 1/3. What is the probability that the marble will be either red or blue?

    (A) 1/6
    (B) 1/4
    (C) 2/7
    (D) 1/3
    (E) 5/12
    Since the problem asks for a FRACTION, the total number of marbles can be ANY VALUE.
    Let the total number of marbles = the LCM of 3 and 4 = 12.

    If one marble is drawn from the box at random, the probability that it will be white is 1/4.
    Thus:
    W = (1/4)(12) = 3.
    If one marble is drawn from the box at random, the probability that it will be green is 1/3.
    Thus:
    G = (1/3)(12) = 4.
    Implication:
    Of the 12 marbles, the number that are R or B = 12-3-4 = 5.

    Since 5 of the 12 marbles are R or B, we get:
    P(R or B) = 5/12.

    The correct answer is E.

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    Post Wed Aug 06, 2014 9:23 pm
    sachin_yadav wrote:
    There is a total of 120 marbles in a box, each of which is red, green, blue, or white. If one marble is drawn from the box at random, the probability that it will be white is 1/4 and the probability that it will be green is 1/3. What is the probability that the marble will be either red or blue?

    (A) 1/6
    (B) 1/4
    (C) 2/7
    (D) 1/3
    (E) 5/12

    OA E

    Please explain

    Regards
    Sachin
    Total Probability = 1
    i.e. 1 = Either White + or Green + or Red + or Blue
    i.e. 1 = (1/4) + (1/3) + (Probability of Red or Blue)
    i.e. (Probability of Red or Blue) = 1 - ((1/4)+(1/3) = 5/12

    Answer: Option E

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    sachin_yadav Really wants to Beat The GMAT!
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    Post Sat Aug 09, 2014 6:28 am
    Thank you for your replies. Appreciate it Very Happy

    Regards
    Sachin

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    Post Sat Aug 09, 2014 12:03 pm
    Hi sachin_yadav,

    Each of the methods outlined in this post can help you to get to the correct answer. Here's one more that is based on converting the data and using the structure of the answer choices to your advantage:

    First off, in almost all cases, when the answers to a GMAT Quant question are numbers, they will be arranged from least to greatest. This will allow you to take advantage of approximation, rounding, etc., so that you can quickly eliminate answers that are "too big" or "too small"

    In this prompt, we're dealing with some simple fractions (1/4 and 1/3), which you should be able to convert to decimal (you should have them memorized):

    1/4 are white = .25 are white
    1/3 are green = .33 are green (approximately)

    .25 + .33 = .58 total of white and green (approximately)

    The question asks for the probability of red or blue, so we have…

    1 - .58 = .42 total of red and blue (approximately)

    The answers here are arranged from least (1/6 = .1666) to greatest (5/12 = let's not mess with this just yet)

    Answer D = 1/3 = .33, which is TOO LOW (we're looking for approximately .42). This means that A, B and C are also TOO LOW.

    Final Answer: E

    GMAT assassins aren't born, they're made,
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