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Black and Yellow Balls

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Black and Yellow Balls

by ronnie1985 » Tue May 22, 2012 10:15 am
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
1/4
1/2
5/8
2/3
3/4
I find this question lengthy. Any shortcut please????
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by Brent@GMATPrepNow » Tue May 22, 2012 11:20 am
The probability that the 4th person selects a black ball is the same as the probability that the 1st person selects a black ball

There are 8 balls, and 5 of them are black, so the probability is [spoiler]5/8 = C[/spoiler]

More here: https://www.beatthegmat.com/probability-t92340.html

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by Anurag@Gurome » Tue May 22, 2012 7:35 pm
ronnie1985 wrote:A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
1/4
1/2
5/8
2/3
3/4
I find this question lengthy. Any shortcut please????
YBBB; probability that 4th ball is black = 3/5 (as we have already picked 2 black balls, so the remaining are 3 balls)
YYBB; probability that 4th ball is black = 4/5 (as we have already picked 1 black ball, so the remaining are 4 balls)
YYYB; probability that 4th ball is black = 5/5 (as we have not picked any black ball, so there are 5 balls)
BBBB; probability that 4th ball is black = 2/5 (as we have already picked 3 black balls, so the remaining are 2 balls)

Therefore required probability = [(3C1 * 5C2)/8C3] * 3/5 + [(3C2 * 5C1)/8C3] * 2/5 + [3C3/8C3] * 1 + (5C3/8C3) * 2/5 = (30/56) * (3/5) + (15/56) * (4/5) + (1/56) + (10/56) * (2/5) = (18 + 12 + 1 + 4)/56 = 35/56 = [spoiler]5/8[/spoiler]

The correct answer is C.
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