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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
Can we use the bionomial method like

8C3(1/4)((1/4)^2) +.... kind??
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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
13/14 =1- 4c3/8c3 (all white) = (4c1x4c2+4c2x4c1+4c3)/8c3 (at least one read)
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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
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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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
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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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
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

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Re: There are 8 marbles in a bag, 4 of them are red and others [#permalink]
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