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How can 20 balls, 10 white and 10 black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it?

Short Answer

Expert verified

The white ball from each urn is the maximizes possibility of drawing is one white ball from urn1and nine white ball and ten black ones from urn2.

Step by step solution

01

Step: 1 Events:

The probability formula is

P(White)=P(WhiteUrn1)P(Urn1)+P(WhiteUrn2)P(Urn2)

Urn randomly as

role="math" localid="1649501479994" P(Urn1)=12P(Urn2)=12P(White)=12[P(WhiteUrn1)+P(WhiteUrn2)]

Since,P(white) is greater than P( white/Urn1) and P(white/Urn 2).

02

Step: 2 Upper bounds:

May be both urns are same as many white and black balls as

P(WhiteUrn1)=P(WhiteUrn2)=12P(White)=1212+12P(White)=12.

One urn less than half white balls.so it's urn 2.

P(WhiteUrn2)<12.

03

Step: 3 Getting probability:

In total 20balls,the nearest probabality can get 1/2is 9/19.So it's upper limit.

Probabilities are less are equal to one.

P(WhiteUrn1)1P(WhiteUrn2)919

It's theoretically maximum and the distribution satisfies maximum.

Urns are not differentiated.if not reverse will be a solution.

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