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There are 5 blue balls and 7 red balls in the 1st box and 11

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There are 5 blue balls and 7 red balls in the 1st box and 11 [#permalink] New post 18 Mar 2005, 11:21
There are 5 blue balls and 7 red balls in the 1st box and 11 blue and 4 red in the 2nd one. Ann takes 1 ball out of a box. What is the probability that the chosen ball is blue? (we don't know out of which box Ann's taken the ball)







Don't have OA. Not sure that numbers are correct. Just faced such a question in one of the tests I have taken.
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 [#permalink] New post 18 Mar 2005, 13:47
The total probability is

P(getting a blue ball from box 1) x P(choosing the box 1) + P(getting a blue ball from box 2) x P(choosing the box 2)

Since there're two boxes (and assuming the boxes are non biased, and identical), the P(choosing box 1) = P(choosing box 2) = 1/2.

Thus total probability =

1/2 x (5/12) + 1/2 x (11/15) = 1/2 x (69/60) = 69/120.

Hope that helps.
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 [#permalink] New post 19 Mar 2005, 11:24
thanks :rotate
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 [#permalink] New post 14 Jul 2005, 00:57
Agree with Capslock. Good Explanation.
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Re: probability [#permalink] New post 15 Jul 2005, 11:09
inga wrote:
There are 5 blue balls and 7 red balls in the 1st box and 11 blue and 4 red in the 2nd one. Ann takes 1 ball out of a box. What is the probability that the chosen ball is blue? (we don't know out of which box Ann's taken the ball)







Don't have OA. Not sure that numbers are correct. Just faced such a question in one of the tests I have taken.


:? What about this approach

plz if i am wrong tell me why

the blue marble can be taken either from the first box or the second

prob of a blue first box =5/12

prob of a blue second box 11/15

5/12+11/5 PROB to have a blue one from either box

thanks :?
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 [#permalink] New post 15 Jul 2005, 16:33
Is the fact that the balls are stored in two seperate boxes affecting the
outcome ? (vs. having them all together)
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 [#permalink] New post 16 Jul 2005, 03:07
My first reaction was that the balls should all be treated as if they were stored together instead of in separate containers. I came up with 16/27 this way. However, Kapslock answer also makes sense - I'm just not sure why you need to use the more complicated formula. Can someone explain why 16/27 isn't correct?
  [#permalink] 16 Jul 2005, 03:07
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There are 5 blue balls and 7 red balls in the 1st box and 11

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