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prob - exactly vs. at least [#permalink]
 08 Jan 2008, 08:34
Please correct me if i am wrong. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that a) exactly one is blue? b) at least one marble is blue? desired / total a) 3C1 * 7C1 / 10C2 = 21/45 b) Total - P(none) / total 10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 08:39
looks good to me 
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 09:04
a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 09:17
bmwhype2 wrote: akhi wrote: a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45 u selected 2 blue. and subtracted incorrectly
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 09:21
eschn3am wrote: bmwhype2 wrote: akhi wrote: a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45 u selected 2 blue. and subtracted incorrectly sorry..my bad..this is wht happens when u try to solve problem at work..  it should be 1-3C2 / 10C2 = 42/45
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 09:23
other ways: a) 1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15 2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15 b) 1. p=1-3/10*2/9=1-1/15=14/15 2. p=1-3P2/10P2=1-3*2/(10*9)=14/15
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 19:00
isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ?
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Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 19:35
pmenon wrote: isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ? Yes, but that's more work than finding finding out the probability of 2 red and just subtracting that from 1.
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 02:12
bmwhype2 wrote: Please correct me if i am wrong.
A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?
desired / total
a) 3C1 * 7C1 / 10C2 = 21/45
Getting 7/15 for a. p(exactly 1 blue) = 7/10 * (1-6/9) * 2! = 7/15
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 03:45
P(one is blue)=(3/10)*(7/10)+(7/10)*(3/10)=21/50*2=7/15
P(at least one is blue)=1-P(red&red)=1-(3/10)*(2/9)=14/15
Or still another way: Probability=#favourable/#total
P(one is blue)=[2!*C(7,1)*C(3,1)]/[10!/8!]=21/45=7/15
Last edited by automan on 09 Jan 2008, 04:05, edited 1 time in total.
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 03:56
automan wrote: P(one is blue)=(3/10)*(7/10)+(7/10)*(3/10)=21/50*2=7/15
P(at least one is blue)=1-P(red&red)=1-(3/10)*(2/9)=14/15 yes, walker is correct
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 06:28
GK_Gmat wrote: bmwhype2 wrote: Please correct me if i am wrong.
A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?
desired / total
a) 3C1 * 7C1 / 10C2 = 21/45
Getting 7/15 for a. p(exactly 1 blue) = 7/10 * (1-6/9) * 2! = 7/15 thats the same value. 21/45 = 7/15
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 08:19
bmwhype2 wrote: Please correct me if i am wrong.
A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?
desired / total
a) 3C1 * 7C1 / 10C2 = 21/45
b) Total - P(none) / total
10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15 A: consider two possiblities: 3/10*7/9 (2) --> 21/45 --> 7/15 b/c we have to consider the scenario of 7/10*3/9. So the multiplied by 2 is the same as 21/90+21/90 --> 42/90 --> 7/15. B: 1-prob. that none will be blue. 3/10*2/9 --> 6/90 --> 15/15-1/15 --> 14/15
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 13:55
Walker is right
for qn1. for exactly one to be blue, it should be BR & RB 7/10*3/8 + 3/10*7/9 = 7/15
for qn2. atleast one Blue, you can add BB to the above & get the answer.
prob of BB is 7/10*6/9 =7/15
so total of BB< RB and BR = 2*7/15= 14/15
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 18:31
Sorry this might just be my ingnorance, but what does "p1... p2" stand for??
a)
1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15
2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15
b)
1. p=1-3/10*2/9=1-1/15=14/15
2. p=1-3P2/10P2=1-3*2/(10*9)=14/15
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Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 23:51
bsjames2 wrote: Sorry this might just be my ingnorance, but what does "p1... p2" stand for??
a)
1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15
2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15
b)
1. p=1-3/10*2/9=1-1/15=14/15
2. p=1-3P2/10P2=1-3*2/(10*9)=14/15 3P1 - permutation formula: 3!/2! nPm=n!/(n-m)!
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Re: prob - exactly vs. at least [#permalink]
10 Jan 2008, 12:02
1. RB + BR = 7/10 * 3/9 + 3/10 * 7/9 = 7/15
2. RB + BR + RR = 7/15 (from above) + 7/10 * 6/9
= 7/15 + 14/30 = 28/30 = 14/15
OR
1 - RR [All cases except when both picks are red]
1 - 3/10 * 2/9 = 1 - 1/15 = 14/15
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Re: prob - exactly vs. at least
[#permalink]
10 Jan 2008, 12:02
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