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There are 8 marbles in a bag, 4 of them are red and others

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Senior Manager
Joined: 10 Dec 2004
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There are 8 marbles in a bag, 4 of them are red and others [#permalink]

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18 May 2005, 09:49
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There are 8 marbles in a bag, 4 of them are red and others are white. If 3 marbles are randomly drawn out from the bag, what is the probability that at least one of the marbles is red?

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Director
Joined: 18 Feb 2005
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18 May 2005, 10:00
Let the 3 things drawn out are white == 4C3 ways

Total no of ways of drawing are 8C3

So probabaility of drawing only white is 4C3/8C3 = 1/14

So probability of red being drawn = 1-1/14 = 13/14

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Manager
Joined: 27 Jul 2004
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18 May 2005, 10:01
1 - Prob(Not picking any RED marble) OR Prob(Picking up all 3 WHITE marbles)

1 - 4C3/8C3

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Director
Joined: 18 Feb 2005
Posts: 668

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18 May 2005, 10:03
Can we use the bionomial method like

8C3(1/4)((1/4)^2) +.... kind??

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Director
Joined: 18 Apr 2005
Posts: 543

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Location: Canuckland

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18 May 2005, 13:34
13/14 =1- 4c3/8c3 (all white) = (4c1x4c2+4c2x4c1+4c3)/8c3 (at least one read)

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Director
Joined: 07 Jun 2004
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Location: PA

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18 May 2005, 13:45
total ways 8C3 = 56

Total ways 3 marbles are white = 4C3 = 4

all white marbles or none red marbles = 4/56 = 1/14

so At least one red = 1 - 1 / 14 = 13/14

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Director
Joined: 20 Apr 2005
Posts: 584

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27 May 2005, 16:12
The chance of at least one red = 1- P(all 3 are white)

= 1- (4/8 x3/7 x2/6)

= 1- 1/14

= 13/14

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Senior Manager
Joined: 21 Mar 2004
Posts: 445

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Location: Cary,NC

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29 May 2005, 11:54
P ( at least 1 red ) = 1 - P( no red )

Also , P ( no red ) = P ( all white ) = 4C3/8C3

Final ans = 1 -4C3/8C3 = 13/14
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

29 May 2005, 11:54
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