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There are three blue marbles, three red marbles, and three yellow marb

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There are three blue marbles, three red marbles, and three yellow marb [#permalink]

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New post 08 Feb 2017, 05:27
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There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one marble of each color from the bowl after three successive marbles are withdrawn from the bowl?

A) 1/27
B) 3/56
C) 3/28
D) 9/56
E) 9/28
[Reveal] Spoiler: OA

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Re: There are three blue marbles, three red marbles, and three yellow marb [#permalink]

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New post 08 Feb 2017, 06:30
There are 9 marbles in the bowl, 3 of each color. The first one that you pick will be the first of that color, regardless of whether it is red, yellow, or blue. Let's say that you select a red marble. On your next selection, you must take a blue or a yellow marble. There are 3 blue marbles and 3 yellow marbles remaining, but there are only 2 red marbles remaining because you already selected one. This mean you have a 6/8 chance of drawing a blue or a yellow. Let's say that you draw a blue. This means that your last marble must be a yellow. There are three yellow marbles remaining but only two red marbles and two blue marbles. This means that you have a 3/7 chance of selecting a yellow marble on your final draw.

This gives us a probability of \(1*\frac{3}{4}*\frac{3}{7}=\frac{9}{28}\). Answer is E
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Re: There are three blue marbles, three red marbles, and three yellow marb [#permalink]

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New post 08 Feb 2017, 06:45
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duahsolo wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one marble of each color from the bowl after three successive marbles are withdrawn from the bowl?

A) 1/27
B) 3/56
C) 3/28
D) 9/56
E) 9/28


P(3 different colors) = P(1st draw is ANY color AND 2nd draw does not match 1st draw AND 3rd draw does not match 1st and 2nd draws)
= P(1st draw is ANY color) x P(2nd draw does not match 1st draw) x P(3rd draw does not match 1st and 2nd draws)
= 1 x 6/8 x 3/7
= 18/56
= 9/28

Answer: E

Cheers,
Brent
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Re: There are three blue marbles, three red marbles, and three yellow marb [#permalink]

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New post 13 Feb 2017, 07:36
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duahsolo wrote:
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one marble of each color from the bowl after three successive marbles are withdrawn from the bowl?

A) 1/27
B) 3/56
C) 3/28
D) 9/56
E) 9/28


We are given that there are three blue marbles, three red marbles, and three yellow marbles in a bowl. We need to determine the probability of selecting exactly one marble of each color from the bowl after three successive marbles are withdrawn from the bowl.

We note that one marble of each color is possible in six ways:

BRY
BYR
YBR
YRB
RBY
RYB

Each of the above scenarios has an equal chance of happening; therefore, we will find the probability that one of them (BRY) will happen and multiply the result by 6.

To draw a blue, red, and yellow marble in this specific order, we first need to draw one of the three blue marbles out of nine marbles; therefore the probability of this event is 3/9 = 1/3. Next, we need to draw one of the three red marbles out of eight remaining marbles (since a blue marble has already been drawn), which has a chance of 3/8. Finally, we need to draw one of the three yellow marbles out of seven remaining marbles (since a blue and a red marble have already been drawn), and this event has a probability of 3/7. Combining the three events, we find that drawing BRY has a probability of 1/3 x 3/8 x 3/7 = 3/56.

Since each of the remaining five events has an equal probability to the event BRY, the probability of drawing a marble of each color is 3/56 x 6 = 9/28.

Answer: E
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Re: There are three blue marbles, three red marbles, and three yellow marb   [#permalink] 13 Feb 2017, 07:36
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