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Probability - breaking the jar ?

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by arora007 » Fri Jul 30, 2010 8:57 am
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).


Good one...
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by nithi_mystics » Fri Jul 30, 2010 9:40 am
b) X/Y

probability of pulling a black ball out of a glass jar and breaking the jar = probability of pulling a black ball out of a glass jar * probability of breaking the jar

1/y = 1/x * probability of breaking the jar

==> probability of breaking the jar = x/y
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by Gurpinder » Fri Jul 30, 2010 9:46 am
arora007 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).


Good one...
to get the probability of something AND something happening you multiply the two probabilities

so that means that 1/x * ______ = 1/y so just divide 1/x / 1/y which equals to x/y

so the answer is B
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by arora007 » Fri Jul 30, 2010 9:56 am
both of u deserve a 51. God bless!!
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by babuxavier » Wed Nov 21, 2012 9:43 am
IMO B
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by Bill@VeritasPrep » Wed Nov 21, 2012 9:45 am
nithi_mystics wrote:b) X/Y

probability of pulling a black ball out of a glass jar and breaking the jar = probability of pulling a black ball out of a glass jar * probability of breaking the jar

1/y = 1/x * probability of breaking the jar

==> probability of breaking the jar = x/y
Yup. The keyword here is AND: black ball AND jar breaks. Nicely done.
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by Jeff@TargetTestPrep » Mon Dec 18, 2017 5:53 pm
arora007 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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