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Thread: Tricky question

  1. #1
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    Tricky question

    I have 9 balls in a bag numbered from 1 to 9. I take 1 ball and I guess its number before looking it. I put it on the side (extraction WITHOUT REPLACEMENT) and I take a second ball from the bag and again a guess its number without looking and I put it on the side. So on for another 3 balls - in total drawing 5 balls. What's the probability that I have guessed 3 out of 5 balls number right? Or better how do I calculate such probability?

    In fact its not a binomial distribution (the probabilities of each event are not independent) or an hypergeometric distribution (since there are not N success and M-N failures)

    Thank you very much!


  2. #2
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    Re: Tricky question

    Let's see. The fact that you're not looking at the balls after you guess the number pretty much rules out the extraction without replacement because since you have no idea the numbers on the balls you've removed, you are still left guessing 1-9 for all remaining balls.

    So really, isn't this actually a binomial distribution? 1 in 9 chance, 5 draws, chance of guessing 3.

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