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# The probability of pulling a black ball out of a glass jar

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Manager
Joined: 14 Dec 2004
Posts: 117
The probability of pulling a black ball out of a glass jar [#permalink]

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13 Mar 2005, 14:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?

a) 1/(XY).
b) X/Y.
c) Y/X.
d) 1/(X+Y).
e) 1/(X-Y).
Manager
Joined: 13 Oct 2004
Posts: 236

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13 Mar 2005, 15:11
P(A And B) = P(A).P(B|A)

Where P(A) = probability of being black
P(B) = Probabilty of breaking the jar

from the data in the prob, 1/y = (1/x) * P(B|A) = x/y.

Not sure if this is the right approach for this problem. As x/y gives the probability of breaking the jar given that a black ball is drawn from the jar.
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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13 Mar 2005, 17:22
B too, same approach as prep_gmat
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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14 Mar 2005, 00:03
Let A denote P(pulling black ball out of glass jar)= 1/x
B denote P(black ball out of glass jar and breaking jar) = 1/y
C denote P(breaking glass jar)

A AND C = B
(1/x)(C) = 1/y
(C) = x/y
Director
Joined: 21 Sep 2004
Posts: 607

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14 Mar 2005, 07:23
P(A and B) = P(A).P(B)
X/Y

Last edited by vprabhala on 14 Mar 2005, 10:56, edited 1 time in total.
VP
Joined: 18 Nov 2004
Posts: 1433

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14 Mar 2005, 09:22
"B"

I think pulling a ball and breaking the jar r independent events so I don't think P(A|B) applies here.

For independent events:

P(A and B) = P(A).P(B)

1/Y = 1/X.P(B)

P(B) = X/Y
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville

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14 Mar 2005, 12:12
X/y
prob of Blakc ball and Breaking jar= Prob of Black ball*prob of breaking jar=1/y
14 Mar 2005, 12:12
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