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16 Oct 2006, 11:42
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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:
45%
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Question Stats:
73% (02:38) correct
27%
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wrong
based on 928
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5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
A. \(\frac{4}{33}\)
B. \((\frac{5}{36})^2\)
C. \(\frac{1}{2}\)
D. \((\frac{31}{36})^2\)
E. \(\frac{29}{33}\)
A. \(\frac{4}{33}\)
B. \((\frac{5}{36})^2\)
C. \(\frac{1}{2}\)
D. \((\frac{31}{36})^2\)
E. \(\frac{29}{33}\)
11 Aug 2010, 08:45
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5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITHOUT replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
Given: 5B+3R+4P=12 marbles. Question: what is the probability of NOT having scenario BBPP.
Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles (scenario BBPP):
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{4}{11}*\frac{4}{10}*\frac{3}{9}=\frac{4}{33}\), we are multiplying by \(\frac{4!}{2!2!}\) as scenario BBPP can occur in several ways: BBPP, PPBB, BPBP, ... So scenario BBPP can occur in \(\frac{4!}{2!2!}\) # of ways (# of permuations of 4 letters BBPP out of which 2 B's and 2 P's are identical);
Or with combinatorics \(P(opposte \ event)=\frac{C^2_5*C^2_4}{C^4_{12}}=\frac{4}{33}\).
So \(P=1-\frac{4}{33}=\frac{29}{33}\).
5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITH replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
Again Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles:
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{5}{12}*\frac{4}{12}*\frac{4}{12}=\frac{25}{216}\), the same reason to multiply by \(\frac{4!}{2!2!}\);
So \(P=1-\frac{25}{216}=\frac{191}{216}\).
Similar problem: probability-colored-balls-55253.html
Hope it helps.
Given: 5B+3R+4P=12 marbles. Question: what is the probability of NOT having scenario BBPP.
Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles (scenario BBPP):
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{4}{11}*\frac{4}{10}*\frac{3}{9}=\frac{4}{33}\), we are multiplying by \(\frac{4!}{2!2!}\) as scenario BBPP can occur in several ways: BBPP, PPBB, BPBP, ... So scenario BBPP can occur in \(\frac{4!}{2!2!}\) # of ways (# of permuations of 4 letters BBPP out of which 2 B's and 2 P's are identical);
Or with combinatorics \(P(opposte \ event)=\frac{C^2_5*C^2_4}{C^4_{12}}=\frac{4}{33}\).
So \(P=1-\frac{4}{33}=\frac{29}{33}\).
5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITH replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
Again Calculate the probability of an opposite event and subtract this value from 1.
Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles:
\(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{5}{12}*\frac{4}{12}*\frac{4}{12}=\frac{25}{216}\), the same reason to multiply by \(\frac{4!}{2!2!}\);
So \(P=1-\frac{25}{216}=\frac{191}{216}\).
Similar problem: probability-colored-balls-55253.html
Hope it helps.
General Discussion
11 Aug 2010, 09:37
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Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?
