You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white. You can place the balls in the urns any way that you'd like, including leaving an urn empty. Your friend will choose one urn at random and then draw a ball from that urn. ( If he chooses an empty urn, he draws nothing.) She wins if she draws the black ball and loses otherwise.
a. Suppose you arrange the balls in the way that minimizes her chances of drawing the black ball. What is her probability of winning?
b.Suppose you arrange the balls in the way that maximizes her chances of drawing the black ball. What is her probability of winning?
c. What are the analogous probabilities when there are n balls total ( one black, n-1 white) instead of 4 balls total?
Let A be the event of choosing an urn 1 and B be the event of choosing black ball.
Then probability of finding a black ball can be found using this formula
There are actually four situations: we will see one by one
1) Consider the situation of drawing a black ball when all the balls are in one urn and other is empty.
Then P(B)=1/4 *1/2 =1/8
2) when urn 1 has the black ball and urn2 has all the three white balls
3) when urn 1 has black ball and one white ball and urn2 has rest of the white balls
4) when urn 1 has black ball and 2 white balls and urn2 has a white ball
a) If all the balls are in one urn her chance of winning is very less ie 1/8..
b) Putting black ball alone in one urn increases her chance of winning with probability 1/2
c)similarly the chance of winning with a maximum probability of 1/2 can be achieved by taking black ball alone in one urn and rest of the n-1 white balls in the other urn
minimum probability can be obtained when all the n balls are taken in one urn
suppose all the balls are taken in urn 1
Then the minimum probability is Probability of getting black ball given urn 1*PRobability of selecting urn 1=(1/n)*(1/2)=1/2n
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