Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

prob of hitting one bin is 1/4, times how many bins = 1/16 times three = 1/48...
Hi 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.
Oh my goodness, you're right!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hi; That is okay, this problem has killing me for a week now. I should have been clearer.
No, 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....
Hi; I have a computer program that gives the exact answer but I can not figure out how to do it.
maybe 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.
computer program huh? that's so cool..
Hi; It is good for getting answers when you need them but sometimes you need the math.
yeah lol
Hi; Thanks for looking at the problem. If you get anything... See ya.
What did you get using your computer program, did you get approx. 4.3 throws?
wost case ide say its 36, suposing there would have to be 2 balls in all 4 bins before one would have 3 in it
With 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.
|dw: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
*throws, lol
The 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 in-between.

Not the answer you are looking for?

Search for more explanations.

Ask your own question