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prob - exactly vs. at least [#permalink]
08 Jan 2008, 07:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 18:00
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 ?
Re: prob - exactly vs. at least [#permalink]
08 Jan 2008, 18:35
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.
Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 01:12
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?
Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 05:28
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 07:19
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
Re: prob - exactly vs. at least [#permalink]
09 Jan 2008, 12:55
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