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Q: Probability calculation ( Answered ,   1 Comment ) Question
 Subject: Probability calculation Category: Science > Math Asked by: kiddgame2000-ga List Price: \$20.00 Posted: 08 Aug 2006 01:52 PDT Expires: 07 Sep 2006 01:52 PDT Question ID: 753739
 ```Recently whilst playing pool i was reaching into the table to lift the balls out so as i could set up the rack. I reached in with both hands at the same time and at the exact same time i lifted out 4 balls in each hand. When i set the balls on the table the 4 balls in my left hand were red and the 4 balls in my right hand were yellow In total there were 15 balls inside the table. 7 were red, 7 were yellow and 1 was black. I want to know what the chances were of this happening, that is the chances of me reaching into the table and lifting out 4 balls of the same colour in one hand and 4 balls of the same colour in the other hand. I also want to see how you work this out as it has been puzzleing me for some time now.``` Subject: Re: Probability calculation Answered By: elmarto-ga on 08 Aug 2006 06:54 PDT Rated: ```Hello! The probability of the event that you take 4 balls of the same colour (any colour) in one hand and 4 balls of the other colour in the other one is about 0.54% The procedure to compute it is the following. Let's first calculate the probability of getting 4 red balls in your left hand and 4 yellow balls in your right hand. When you pick up the first ball with your left hand, there is a 7/15 probability that you pick up a red ball. If you do pick up a red ball, then there will be a 6/14 probability (6 red balls, over 14 balls left) that you pick up another red ball. Again, if you pick a red ball, then there is a 5/13 probability that the next one will be red once again, and for the final ball with your left hand there is a 4/12 probability. Therefore, the probability of picking 4 red balls with your left hand is: (7/15)*(6/14)*(5/13)*(4/12) = 0.0256... Now let's continue with the right hand. There are now 11 balls left in the table. Therefore, the probability of picking a yellow ball is 7/11. If you pick a yellow ball first, then there is a 6/10 probability of picking another yellow one. Following the same logic as above, the next probabilities are 5/9 and 4/8. Therefore, the probability of picking 4 yellow balls in your right hand, given that you picked 4 red ones in your left hand is: (7/11)*(6/10)*(5/9)*(4/8) We thus conclude that the probability of picking 4 red balls in your left hand and 4 right balls in your right hand is: (7/15)*(6/14)*(5/13)*(4/12)*(7/11)*(6/10)*(5/9)*(4/8) = 0.0027 However, you asked for the probability of picking 4 balls of any one color in one hand and 4 of the other color in the other one. Therefore, we must also take in account the probability that you get 4 yellow balls in your left hand and 4 red balls in your right hand. Obviously, this probability is exactly the same: 0.0027. We thus get that the probability of picking 4 balls of any one color in one hand and 4 of the other color in the other one is 0.0027+0.0027 = 0.0054 (or 0.54%) I hope this helps! If you have any doubt regarding my answer, please don't hesitate to request clarification before rating it. Otherwise, I await your rating and final comments. Best wishes! elmarto```
 kiddgame2000-ga rated this answer: ```Thats excellent thank you very much, this has been bugging me for quite a while now. Thank you``` `Thank you for the nice comments and rating!` 