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08 Feb 2017, 06:27
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (01:54) correct
50%
(01:49)
wrong
based on 171
sessions
History
Date
Time
Result
Not Attempted Yet
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What is the probability of selecting exactly one marble of each color from the bowl after three successive marbles are withdrawn from the bowl?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
A) 1/27
B) 3/56
C) 3/28
D) 9/56
E) 9/28
08 Feb 2017, 07:30
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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
This gives us a probability of \(1*\frac{3}{4}*\frac{3}{7}=\frac{9}{28}\). Answer is E
08 Feb 2017, 07:45
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duahsolo
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
