A jar contains only x black balls and y white balls

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uptowngirl92: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 19, 2009 9:08 pm; Location: Kolkata,India; Thanked: 7 times; GMAT Score:670

A jar contains only x black balls and y white balls

by uptowngirl92 » Sat Jun 06, 2009 4:04 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) [x/(x+y)] [y/(x+y)]

(B)[x/(x+y)] [(x-1)/(x+y-1)]

(C) xy/x+y

(D) [(x-1)/(x+y)] [(y-1)/(x+y)]

(E)[x/(x+y)] [y/(x+y-1)]

[spoiler]OA:E[/spoiler]

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Source: Beat The GMAT — Problem Solving | Sat Jun 06, 2009 4:04 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 am; Thanked: 712 times; Followed by:550 members; GMAT Score:770

by DanaJ » Sat Jun 06, 2009 5:36 am

Well the probability of an event is:
number of favorable cases
------------------------------
number of possible cases

For your first draw, you're looking to get a black ball. There are (x + y) balls to choose from, but only x are favorable (cause you have only x black balls), with probability x/(x + y).

For your second draw, you have (x + y - 1) balls left, since you've already extracted one ball. You're trying to get a white ball, and there are y white balls left if you consider that your first draw was successful. In this case, the probability will be y/(x + y - 1).

The probability of both events will be obtained by multiplying the two probabilities above.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed Jul 25, 2018 4:40 pm

uptowngirl92 wrote: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) [x/(x+y)] [y/(x+y)]

(B)[x/(x+y)] [(x-1)/(x+y-1)]

(C) xy/x+y

(D) [(x-1)/(x+y)] [(y-1)/(x+y)]

(E)[x/(x+y)] [y/(x+y-1)]

The total number of balls in the jar is (x + y). The probability of drawing a black ball followed by a white ball, without replacement, is:

x/(x + y) * y/(x + y - 1)

Answer: E

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