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?
There are three blue marbles, three red marbles ....
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- EconomistGMATTutor
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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.
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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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.
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.
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We are given that there are three blue marbles, three red marbles, and three yellow marbles in a bowl and need to determine the probability of drawing a yellow marble on each of three draws.Vincen wrote: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?
On the first draw, since there are 3 yellow marbles and 9 total marbles, there is a 3/9 chance that a yellow marble will be selected. Next, since there are 2 yellow marbles and 8 total marbles left, there is a 2/8 chance a yellow marble will be selected. Finally, since there is 1 yellow marble and 7 total marbles left, there is a 1/7 chance that a yellow marble will be selected.
Thus, the total probability is:
3/9 x 2/8 x 1/7 = 1/3 x 1/4 x 1/7 = 1/84
Answer: C
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