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

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The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 17 Dec 2010, 05:39
2
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

70% (01:04) correct 30% (01:31) wrong based on 157 sessions

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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)
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The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 17 Dec 2010, 05:48
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anilnandyala wrote:
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).



Probability of A and B:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: \(P(A \ and \ B) = P(A)*P(B)\)

P(break & black) = P(break) * P(black) --> 1/y=P(break)*1/x --> P(break)=x/y.

Answer: B.
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Re: The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 25 May 2016, 08:24
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anilnandyala wrote:
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)


P of pulling a black ball out of a glass jar AND breaking the jar = 1/Y
P of pulling a black ball out of a glass jar is 1/X
Lets say P of breaking the jar = n

That means 1/X*n=1/Y
n= X/Y

B is the answer
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Re: The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 18 Dec 2010, 00:38
pretty new to this forum and have a question. do you not use plug in or substitution and try to use just the math specifically?

If substitute here, much easier, i think. 1/x substitute x=5 1/y = substitute y = 6 (it has to be less likely to break and black ball, otherwise, there wouldnt be a black ball in the jar) then you have...

1/5 * P(both) = 1/6

5(1/5 * P(both)) = (1/6)5

P(both) = 5/6
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Re: The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 20 Jul 2017, 16:49
anilnandyala wrote:
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)


We can let the probability of breaking the jar = p; thus:

(1/X)p = (1/Y)

p = (1/Y)/(1/X)

p = X/Y

Answer: B
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Re: The probability of pulling a black ball out of a glass jar  [#permalink]

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New post 12 Sep 2019, 23:16
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Re: The probability of pulling a black ball out of a glass jar   [#permalink] 12 Sep 2019, 23:16
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