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  A jar contains only x black balls and y white balls [#permalink]
New postPosted: Sat Jun 06, 2009 4:54 am 
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A jar contains only x black balls and y white balls. One ball is drawn randomly from the jar and is not replaced. A second ball is then drawn randomly from the jar. What is the probability that the first ball drawn is black and the second ball drawn is white?

A. \frac{x}{x+y}*\frac{y}{x+y}

B. \frac{x}{x+y}*\frac{x-1}{x+y-1}

C. \frac{xy}{x+y}

D. \frac{x-1}{x+y}*\frac{y-1}{x+y}

E. \frac{x}{x+y}*\frac{y}{x+y-1}
[Reveal] Spoiler: OA
E

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  Re: A jar contains only x black balls and y white balls [#permalink]
New postPosted: Sat Jun 06, 2009 9:09 am 
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Black balls = x
White balls = y

Total no of balls = x + y

Probability of first ball drawn is black = x/(x+y)
As the ball is not replaced now the total remaining balls = (x+y-1)

Probability of second ball drawn is white = y/(x+y-1)

Total probability = [\frac{x}{(x+y)}]*[\frac{y}{(x+y-1)}]


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  Re: A jar contains only x black balls and y white balls [#permalink]
New postPosted: Tue Jan 10, 2012 7:18 am 
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From 1000ps section 21-13.

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  Re: A jar contains only x black balls and y white balls [#permalink]
New postPosted: Tue Feb 21, 2012 2:02 pm 
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It seems difficult, but actually is one of the easiest.
Just do it.

\frac{x}{(x+y)} * \frac{y}{(x+y-1)}

+1 E

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