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A jar contains only x black balls and y white balls. One ball is drawn

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Bunuel
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A jar contains only x black balls and y white balls. One ball is drawn [#permalink] New post   24 Jul 2018, 01:28
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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. \(\frac{x}{x+y}*\frac{y}{x+y-1}\)
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E
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Re: A jar contains only x black balls and y white balls. One ball is drawn [#permalink] New post   24 Jul 2018, 01:41
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Solution



Given:
    • 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

To find:
    • The probability that the first ball drawn is black and the second ball drawn is white

Approach and Working:
    • The probability that the first drawn ball is black = \(\frac{x}{x+y}\)

As we are not replacing the ball, when the second ball is drawn, the number of balls left is x + y – 1
    • Therefore, the probability that the second drawn ball is white = \(\frac{y}{x+y-1}\)

    • So, the final probability = \(\frac{x}{x+y} * \frac{y}{x+y-1}\)

Hence, the correct answer is option E.

Answer: E
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A jar contains only x black balls and y white balls. One ball is drawn [#permalink] New post   24 Jul 2018, 02:08
Bunuel 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. \(\frac{x}{x+y}*\frac{y}{x+y}\)

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



Note : We are looking for AND probability. Multiplication must be there.

2 types of ball = black and white.

No of black balls = x

No of white balls = y

Total = x + y

probability of getting a black ball ( 1st ) = \(\frac{x}{x+y}\)

Balls left = x + y -1

Probability of getting white ball = \(\frac{y}{x+y -1}\)

Finally multiply both the probability in order to get the result.

Option E is the best answer.
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Re: A jar contains only x black balls and y white balls. One ball is drawn [#permalink] New post   25 Jul 2018, 17:47
Expert Reply
Bunuel 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. \(\frac{x}{x+y}*\frac{y}{x+y}\)

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


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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Re: A jar contains only x black balls and y white balls. One ball is drawn [#permalink] New post   04 May 2023, 04:07
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Re: A jar contains only x black balls and y white balls. One ball is drawn [#permalink]
04 May 2023, 04:07
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