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We are not here to solve problems for you. Mursalin H. told you how to solve the first problem, and if you understand it completely, you should be able to solve this one yourself. I have also provided you with the Brilliant post about the expected value, which will help you understand the concept.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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2^{34}
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Comments
The answer is 13. [provided that all the balls have an equal chance of being selected]
The expected value of a random variable is the weighted average of all possible outcomes.
In your example the probability of a ball being selected is 251. So the expected value would be:
1(251)+2(251)+3(251)+4(251)+⋯+23(251)+24(251)+25(251)
=13.
Hope this helps!
You might want to read this: Expected Value | Brilliant
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Something's terribly wrong with your link. It redirects me to this page again.
You probably wanted to post this link.
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Thanks for pointing that out, I've updated it.
If I choose a ball again with out replace the first ball .What is the expected value of the second ball?
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We are not here to solve problems for you. Mursalin H. told you how to solve the first problem, and if you understand it completely, you should be able to solve this one yourself. I have also provided you with the Brilliant post about the expected value, which will help you understand the concept.
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Just call me Mursalin. Mursalin H. sounds really odd! :)
wen all are equiprobable how can one of them(13) have more chances of appearing