- #1
akoska
- 22
- 0
Urgent probability questions!
Please help!
How can 5 black and 5 yellow balls be put into two urns to maximize the probability that a yellow ball is drawn from a randomly chosen urn?
I got:
P(draw yellow)=P(draw yellow intersect urn1)+P(draw yellow intersect urn2)
=P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2)
=1/2((y/y+k)+(5-y)/(10-y-k))
where y=number of yelow balls in urn 1 and k=number of black balls in urn 2
and y, k<=5
Then, I think I should plug in k=1, 2, 3..., differentiate the above, and set it to zero. Find critical value--but I never get a solution! I get something like -9=0...
Also, if there are b black balls and y yellow balls in an urn, and a person takes one out, replaces it, and adds in another of the same color, prove that at the nth trial, the probability of picking a black ball is b/b+y
Please help!
How can 5 black and 5 yellow balls be put into two urns to maximize the probability that a yellow ball is drawn from a randomly chosen urn?
I got:
P(draw yellow)=P(draw yellow intersect urn1)+P(draw yellow intersect urn2)
=P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2)
=1/2((y/y+k)+(5-y)/(10-y-k))
where y=number of yelow balls in urn 1 and k=number of black balls in urn 2
and y, k<=5
Then, I think I should plug in k=1, 2, 3..., differentiate the above, and set it to zero. Find critical value--but I never get a solution! I get something like -9=0...
Also, if there are b black balls and y yellow balls in an urn, and a person takes one out, replaces it, and adds in another of the same color, prove that at the nth trial, the probability of picking a black ball is b/b+y