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There are three blue marbles, three red marbles ....

This topic has 2 expert replies and 0 member replies

There are three blue marbles, three red marbles ....

Post Tue Dec 05, 2017 9:48 am
There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting a yellow marble from the bowl on each of three successive draws from the bowl?

A. 2/243
B. 1/27
C. 1/84
D. 2/9
E. 5/21

The OA is C.

What are the formulas I should use here? I can find the correct answer. Experts, may you give me some help please?

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Post Tue Dec 05, 2017 6:36 pm
This isn't really answerable because it isn't clear if we're replacing the marbles between draws. (It *sounds* like we aren't, but it isn't clear.)

If we ARE replacing the marbles, then it's just ((# of yellow marbles) / (# of marbles))³

If we AREN'T replacing the marbles, then it's (# of yellow / total) * ((# of yellow - 1)/(total - 1)) * ((# of yellow - 2)/(total - 2)

So the answer's either 1/27 (replacing) or 1/84 (not replacing). Since both answers are there, I'd say you can burn whatever book this came from. Very Happy

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Post Tue Dec 05, 2017 10:49 am
Hello Vincen.

Let's see what are the formulas here.

We have to pick 1 yellow marble from 3 yellow marbles among 9 total marbles. 3/9=1/3.

Now, we have to pick another yellow marble from 2 yellow marbles among 8 total marbles. 2/8=1/4.

Finally, we have to pick another yellow marble from 1 yellow marble among 7 total marbles. 1/7.

So, the final probability is $$\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{7}=\frac{1}{84}.$$ So, the correct answer is C.

I really hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.

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