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rxs0005
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 31 Aug 2010, 13:23
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25% (01:03) correct 75% (01:31) wrong based on 11 sessions

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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards are drawn with replacement from the deck. What is the probability that all 3 cards are different colors


2/9
3/20
1/5
2/7
8/243

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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 31 Aug 2010, 14:15
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2/9*3/9*4/9 = 8/243

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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 31 Aug 2010, 15:17
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rxs0005
A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards are drawn with replacement from the deck. What is the probability that all 3 cards are different colors


2/9
3/20
1/5
2/7
8/243
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1- (RRR+BBB+GGG)

\(1- {(\frac{2}{9}*\frac{2}{9}*\frac{2}{9})+(\frac{3}{9}*\frac{3}{9}*\frac{3}{9})+(\frac{4}{9}*\frac{4}{9}*\frac{4}{9}} = \frac{8+27+64}{9} = \frac{99}{729} = \frac{11}{81}\)
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 31 Aug 2010, 18:53
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anshumishra
2/9*3/9*4/9 = 8/243

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Why are you not multiplying the probability with 3! ? Since the different ways to draw three different cards is 3!.

RGB, RBG, GRB, GBR, BRG and BGR.

Hence shouldn't the answer be 3! * (8/243) --(16/81).
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 31 Aug 2010, 22:08
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It should be 16/81. Why not?
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar Updated on: 02 Sep 2010, 14:57
Originally posted by anshumishra on 01 Sep 2010, 04:53.
Last edited by anshumishra on 02 Sep 2010, 14:57, edited 1 time in total.
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Yeah, I agree it should be 16/81.
I missed to multiply by 3!

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rxs0005
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 01 Sep 2010, 09:04
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The OA for this is 2/7
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Bunuel
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 01 Sep 2010, 09:23
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rxs0005
A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards are drawn with replacement from the deck. What is the probability that all 3 cards are different colors


2/9
3/20
1/5
2/7
8/243
Show more
rxs0005
The OA for this is 2/7
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This is poor quality question, with bad wording and wrong OA.

We are looking for the probability of RBG.

If the drawing is WITH replacement (as in the original question) then: \(P(RBG)=3!*\frac{2}{9}*\frac{3}{9}*\frac{4}{9}=\frac{16}{81}\), multiplying by 3! as scenario RBG can happen in 3! # of ways (RBG, RGB, ...);

If the drawing is WITHOUT replacement then \(P(RBG)=3!*\frac{2}{9}*\frac{3}{8}*\frac{4}{7}=\frac{2}{7}\), the same reason of multiplying by 3!.

If the question is posted exactly as it is in the source and the OA is indeed 2/7 then I wouldn't study from this source. Also I suggest not to spend time on studying, posting, answering bad quality questions.
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A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 01 Sep 2010, 23:51
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Agree with Bunuel.. This answer made me unlearn the 3! Approach...

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