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17 Dec 2010, 05:39
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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)
A. 1/(XY)
B. X/Y
C. Y/X
D. 1/(X+Y)
E. 1/(X-Y)
17 Dec 2010, 05:48
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anilnandyala
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).
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.
General Discussion
18 Dec 2010, 00:38
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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
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
