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anonymous
 3 years ago
Hi;
There are 4 bins. I throw balls at them until there are 3 balls in one of the bins. Then I stop. The probability of hitting a bin are all the same, 1 / 4. What is the expected number of throws?
anonymous
 3 years ago
Hi; There are 4 bins. I throw balls at them until there are 3 balls in one of the bins. Then I stop. The probability of hitting a bin are all the same, 1 / 4. What is the expected number of throws?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0prob of hitting one bin is 1/4, times how many bins = 1/16 times three = 1/48...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi rajathsbhat; Thanks for replying but 64 throws is too high. You can not distribute more than 9 balls in the 4 bins without there being 3 in one bin. This implies that that 9 throws is the maximum. Of course you can do it in a minimum of 3 throws. So the expected number of throws is somewhere in between 3 and 9. I should have been clearer about the fact that the balls stay in the bins after each throw.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh my goodness, you're right!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi; That is okay, this problem has killing me for a week now. I should have been clearer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, you couldn't have have been any clearer. I mean, if the balls were removed as soon as they fell into a bin, there is no way that you can end up with three balls in one bin...i wonder what i was thinking when i was typing that reply....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi; I have a computer program that gives the exact answer but I can not figure out how to do it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe if the question was something like "what is the probability of getting three balls into one bin in consecutive throws, regardless of how many balls are in the bin already", my answer would be right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0computer program huh? that's so cool..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi; It is good for getting answers when you need them but sometimes you need the math.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi; Thanks for looking at the problem. If you get anything... See ya.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0What did you get using your computer program, did you get approx. 4.3 throws?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wost case ide say its 36, suposing there would have to be 2 balls in all 4 bins before one would have 3 in it

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0With 9 balls, it is a certainty.But they need to fill only any one bin, so we don't need to multiply by 4 to get 36.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356470323583:dw it takes 4 thoughts to gaurentee that there will be a ball in a given bin, that times 4 means that you can expect, worst case to have a ball in each bin by the time you throw 16 balls, times 2 for 2 balls in each bin and you have 32, then + four to make sure that one has at least 3

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.0The worst case is 9 throws, as you have illustrated, 2 in each bin, and the 9th can land in any bin to get 3 balls. The best case is 3 all in the same bin. The answer is somewhere inbetween.
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