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This is from my math book: 427) A box contains 1 black ball

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This is from my math book: 427) A box contains 1 black ball [#permalink] New post 28 Feb 2005, 11:55
This is from my math book:

427) A box contains 1 black ball and 9 white balls. Another box contains x black balls and the remaining balls are white, with a total of 10 balls. One ball is picked randomly from the first box and put into the second box, but its colour is not registered. Then one ball is picked from the second box. The probability that this last ball is black is:

(A) x/10 + 1/10
(B) x/11 + 1/10
(C) x + 1/11
(D) x/10 + 1/110
(E) x/11 + 1/110

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 [#permalink] New post 28 Feb 2005, 12:07
"E"


2 scenarios:

1 B is picked from 1st basket, then then placed into second, so p(e) in that case = (1/10)*(x+1)/11

1 W is picked from 1st basket, then then placed into second, so p(e) in that case = (9/10)*x/11


Total = (x+1)/110 + 9x/110 = (10x + 1)/110 = x/11 + 1/110
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 [#permalink] New post 28 Feb 2005, 13:15
E) Did the same way as Banerjee
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 [#permalink] New post 28 Feb 2005, 14:16
The Probab = Prob of picking black from 1st and then from 2nd too + Probab of picking white from 1st and then white from the 2nd

P = 1/10 * (x+1)/11 + 9/10 * x/11 = (x+1+9x)/110 = x/11 + 1/110
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 [#permalink] New post 02 Mar 2005, 07:26
Yup, this is it!

OA is (E).
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 [#permalink] New post 02 Mar 2005, 08:29
E ...sorry was late with this one
  [#permalink] 02 Mar 2005, 08:29
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This is from my math book: 427) A box contains 1 black ball

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