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

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Intern
Joined: 04 Feb 2005
Posts: 11
This is from my math book: 427) A box contains 1 black ball [#permalink]

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28 Feb 2005, 12:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Regards.
VP
Joined: 18 Nov 2004
Posts: 1433

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28 Feb 2005, 13:07
"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
Director
Joined: 19 Nov 2004
Posts: 556
Location: SF Bay Area, USA

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28 Feb 2005, 14:15
E) Did the same way as Banerjee
Manager
Joined: 24 Jan 2005
Posts: 217
Location: Boston

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28 Feb 2005, 15:16
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
Intern
Joined: 04 Feb 2005
Posts: 11

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02 Mar 2005, 08:26
Yup, this is it!

OA is (E).
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville

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02 Mar 2005, 09:29
E ...sorry was late with this one
02 Mar 2005, 09:29
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