ydmuley wrote:
Hello
Bunuel and
VeritasPrepKarishmaAs per
Veritas the answer is 2/7
And the logical argument is that event is happening simultaneously hence we will only consider one scenario of removing the balls.
Could you please confirm what should be the correct strategy and what is the correct answer?
We are now confused between 4/7 & 2/7
Bunuel wrote:
Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same. The correct answer is 4/7.
Ok.. thanks
Bunuel - will go with 4/7
VeritasPrepKarishma - Not sure if you agree to this, in case the books have to be changed to avoid confusion going forward.
Responding to a pm:
The answer is certainly 4/7 and that is what the book says too.
The book shows that "2 simultaneous picks" is the same as "pick one and then another". So getting one black and one white can be achieved in 2 ways: a black and then a white or a white and then a black. I agree that the addition isn't explicitly shown but "pick two such that one is black and one is white" is composed of two cases:
First black and then white for which Probability = 2/7
First white and then black for which Probability = 2/7
They both result in one white and one black so answer would be 4/7.
Note that there are only 2 other cases:
Both black for which probability = 4/7 * 3/6 = 2/7
Both white for which probability = 3/7 * 2/6 = 1/7
Overall probability = 2/7 + 2/7 + 2/7 + 1/7 = 1
_________________
Karishma
Veritas Prep GMAT Instructor
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