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14 Apr 2007, 07:51
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Hi All,
For following question I am getting two different questions by two different methods. can some please clarify the what's the mistake i am making.
There are 2 red and 6 blue balls in a bag. If a person takes out 2 balls at random, what is the probability that at least one of the balls is red?
Apporach 1:
Finding the probability that none of the balls are red = 1-both balls are Blue
1-(6/8)*(5/7)= 1- (30/56) =26/56 =13/28
Apporach 2: Finding teh cases when 1 or two red balls are withdrawn.
(2/8)(6/7) + (2/8)*(1/7) =(12/56) + (2/56) =(14/56) =(7/28)
So I am missing something please point out the mistake and provide the solution to the question
For following question I am getting two different questions by two different methods. can some please clarify the what's the mistake i am making.
There are 2 red and 6 blue balls in a bag. If a person takes out 2 balls at random, what is the probability that at least one of the balls is red?
Apporach 1:
Finding the probability that none of the balls are red = 1-both balls are Blue
1-(6/8)*(5/7)= 1- (30/56) =26/56 =13/28
Apporach 2: Finding teh cases when 1 or two red balls are withdrawn.
(2/8)(6/7) + (2/8)*(1/7) =(12/56) + (2/56) =(14/56) =(7/28)
So I am missing something please point out the mistake and provide the solution to the question
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14 Apr 2007, 08:21
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Text-book way of solving:
Since the problem is asking for "at least one" it's quicker to use formula,
P(at least one) = 1 - P(none)
So u find out the prob of drawing only 2 blue balls (all blue).
P(2 all blue) = (6/8)(5/7) = 30/56
Therefore, 1-(30/56)=13/28.
I think with your approach 2, you should've used combinations. Cuz you don't know "which" of the two red balls and six blue balls were chosen. You have to take into account red1 chosen with blue1, red1 chosen with blue 2 etc.
There are 2*6=12 ways of picking 1 red and 1 blue.
There is only ONE way of picking 2 red. (There are only 2 reds!!)
So total, 13.
Now, you have 8C2=28 ways of picking two balls from a bag of eight.
Thus, 13/28.
Since the problem is asking for "at least one" it's quicker to use formula,
P(at least one) = 1 - P(none)
So u find out the prob of drawing only 2 blue balls (all blue).
P(2 all blue) = (6/8)(5/7) = 30/56
Therefore, 1-(30/56)=13/28.
I think with your approach 2, you should've used combinations. Cuz you don't know "which" of the two red balls and six blue balls were chosen. You have to take into account red1 chosen with blue1, red1 chosen with blue 2 etc.
There are 2*6=12 ways of picking 1 red and 1 blue.
There is only ONE way of picking 2 red. (There are only 2 reds!!)
So total, 13.
Now, you have 8C2=28 ways of picking two balls from a bag of eight.
Thus, 13/28.
14 Apr 2007, 09:58
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ricokevin
Hi Kevin,
Thanls for your reply but the method you have applied R1B1, R2B1 etc why can't that be applied in the 6/8*5/8 apporach? Do you see what I mean...B1B2, B1B3, B1B4,B1B5 etc.
Thanks
Saurabh Malpani
