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 March 29th, 2018, 08:55 AM #1 Newbie   Joined: Mar 2018 From: Ohio Posts: 1 Thanks: 0 average number A bag contains 5 white balls. The following process is repeated. A ball is drawn uniformly at random from the bag (that is each of the five balls have equal probability (= 1/5 ) of being drawn in each trial). If the color of the drawn ball is white then it is colored with black and put into the bag. If the color of the drawn ball is black then it is put into the bag without changing its color. What is the expected average number of times (up to two decimal places) the process has to berepeated so that the bag contains only black balls.
April 6th, 2018, 08:35 AM   #2
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Quote:
 Originally Posted by Mark Evans What is the expected average number of times (up to two decimal places) the process has to be repeated so that the bag contains only black balls.
Five different bags, and moving from one to the next requires a success (white draw); how long until entering the last (all-black) bag? Calculate the success at each bag. More specifically, how long do I gotta draw from a bag when I know the stats in order to get a success and move on?

The first bag is guaranteed 5/5 outcomes lead to the next bag. The second bag will succeed 4/5 of the time, the third 3/5, the forth 2/5 and the fifth 1/5th of the draws will be white.

So the next question is: Given a draw that succeeds 4/5 of the time, how many draws do I need to guarantee a success? Guarantee is a level of confidence in the answer, a certainty, a willingness to be wrong rarely in exchange for narrowing down the prediction. So this part of the problem is to decide how much of a guarantee, how certain in the result we wish to be. The typical value used is 19/20 (95%) successes/tries.

Last edited by AngleWyrm2; April 6th, 2018 at 08:40 AM.

 April 9th, 2018, 04:34 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 I don't think confidence intervals or percentages are used for expected values. If each ball needs to be chosen at least once, the simple expected value formula is: X/X + X/(X-1) + X/(X-2) + ... X/1 For 5 balls, it is 5/5 + 5/4 + 5/3 + 5/2 + 5/1 = 11 5/12 = 11.41(6 repeating)
April 9th, 2018, 04:36 PM   #4
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Quote:
 Originally Posted by EvanJ 11.41(6 repeating)
of course

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