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OG2016 DS A box contains only

This topic has 3 expert replies and 0 member replies
lionsshare Senior | Next Rank: 100 Posts
Joined
09 Aug 2017
Posted:
62 messages

OG2016 DS A box contains only

Post Mon Aug 28, 2017 3:47 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will be either white or blue?

    (1) The probability that the chip will be blue is 1/5.
    (2) The probability that the chip will be red is 1/3.

    OA: B

    Hi, Experts. Please, anyone, explain the solution for this problem. Thank you.

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    Post Tue Aug 29, 2017 8:20 am
    If the box contains only red, white, and blue chips:
    (probability of selecting red) + (probability of selecting white) + (probability of selecting blue) = 1

    The questions asks us for the probability of white or blue chips, so we can treat the group (white or blue) as simply (non-red).
    (probability of selecting red) + (probability of selecting NON-red) = 1

    Thus, if we know the probability of selecting a red, we'll know the probability of selecting anything else (white or blue).

    Target question: what is the probability of selecting red?

    (1) The probability that the chip will be blue is 1/5.
    This gives us blue alone, but not white. Insufficient.

    (2) The probability that the chip will be red is 1/3.
    Sufficient. This answers our target question. If we know that the probability of selecting red is 1/3, then the probability of anything else (white or blue) must be 2/3.

    The answer is B.

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    Post Tue Aug 29, 2017 10:30 pm
    lionsshare wrote:
    A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will be either white or blue?

    (1) The probability that the chip will be blue is 1/5.
    (2) The probability that the chip will be red is 1/3.

    OA: B

    Hi, Experts. Please, anyone, explain the solution for this problem. Thank you.
    Since the box contains only red, white, and blue chips,

    (probability of selecting a red chip) + (probability of selecting a white chip) + (probability of selecting a blue chip) = 1

    Say probability of selecting a red chip = r, probability of selecting a white chip = w, probability of selecting a blue chip = b

    Thus, r + w + b = 1

    We have to find out the probability that the chip will be either white or blue.

    Or, we have to find out the value of (w + b).

    Statement 1: The probability that the chip will be blue is 1/5.

    => b = 1/5. But we do not know the value of w. Insufficient.

    Statement 2: The probability that the chip will be red is 1/3.

    => r = 1/3

    => w + b = 1 - r = 1 - 1/3 = 2/3. Sufficient.

    The correct answer: B

    Hope this helps!

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    Post Thu Aug 31, 2017 5:13 pm
    The trick here:

    P(W) + P(B) + P(R) = 1

    We want P(W) + P(B). Notice that by subtracting P(R) from both sides of the equation above, we have

    P(W) + P(B) = 1 - P(R)

    So knowing P(R) will give us P(W) + P(B)! Once we're told in S2 that P(R) = 1/3, we've got

    P(W) + P(B) = 1 - 1/3

    P(W) + P(B) = 2/3

    and we're set!

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