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13 Jun 2022, 08:15
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D
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Difficulty:
45%
(medium)
Question Stats:
58% (01:01) correct
42%
(01:14)
wrong
based on 250
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A box contains 2 black balls and 2 white balls. If 2 balls are randomly selected from the box, without replacement, what is the probability the two balls are different colors?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4
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13 Jun 2022, 08:49
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BrentGMATPrepNow
Because we do not replace the marble between selections, this is a TWO-STEP probability question. We can NOT simply count the NUMBER of ways to "win," count the total NUMBER of possible ways, and then divide. We need to calculate the PROBABILITIES.
In probability, AND means multiply and OR means add.
BB, BW, WB, WW are all of the possible branches of the probability tree. We can either calculate the probability of "winning" or calculate the probability of losing and subtract that from 1. Either way, there are two branches to calculate, so let's just do "win."
BW: We need B AND then white, so we will multiply. The probability of picking black first is 1/2. We are left with 1 black and 2 white, so the probability of picking a white second is 2/3. 1/2 * 2/3 = 2/6 = 1/3
WB: We should be able to see that this is symmetrical to BW. 1/3
We need either BW OR WB, so we will now add. 1/3 + 1/3 = 2/3
Answer choice D.
15 Jun 2022, 06:48
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BrentGMATPrepNow
P(different colors) = P(1st ball is white AND 2nd ball is black OR 1st ball is black AND 2nd ball is white)
= [P(1st ball is white) x P(2nd ball is black)] + [P(1st ball is black) x P(2nd ball is white)]
= [2/4 x 2/3] + [2/4 x 2/3]
= 1/3 + 1/3
= 2/3
Answer: D
