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19 Dec 2016, 09:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:14) correct
24%
(01:36)
wrong
based on 203
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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 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
A. 2/243
B. 1/27
C. 1/84
D. 2/9
E. 5/21
19 Dec 2016, 10:36
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\(\frac{3}{9} * \frac{2}{8} * \frac{1}{7} = \frac{1}{3} * \frac{1}{4} * \frac{1}{7} = \frac{1}{84}\)
or
\(\frac{3C3}{9C3} = \frac{1}{84}\)
Answer C
or
\(\frac{3C3}{9C3} = \frac{1}{84}\)
Answer C
03 Jan 2017, 17:48
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Bunuel
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.
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
