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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OG2016 DS A box contains only
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lionsshare
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ceilidh.erickson
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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.
(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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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Jay@ManhattanReview
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Since the box contains only red, white, and blue chips,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.
(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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Matt@VeritasPrep
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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!
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!
