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B Random Variable - Mean and Variance

  1. Nov 7, 2018 #1
    Problem:

    We play roulette in a casino. We watch 100 rounds that result in a number between 1 and 36. and count the number of rounds for which the result is odd.

    assuming that the roulette is fair, calculate the mean and deviation

    Solution:

    I understand that the probability - Pr = 0.5. and mean - E[x] = 50. However, I cant figure out how to write it appropriately (use formula E[x] = ∑ x * Pr)

    As I Understand, I must choose random variable x = 1, but don't quite understand the "true" meaning of it and why I can't choose other variable.

    Moreover, I see from solutions that σs=√(100 * 0.5 * 0.5) = 5. However, I cant understand how we get this
     
  2. jcsd
  3. Nov 7, 2018 #2

    andrewkirk

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    The number of odd results observed is a binomial random variable. Have you learned about those? If not, the wikipedia article explains them and gives formulas for the mean and standard deviation.
     
  4. Nov 9, 2018 #3

    Stephen Tashi

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    The number 0.5 is the probability of the event that a single round will be odd. The question asks about different events. It asks about "the number of rounds for which the result is odd" in the 100 rounds. Letting X be that random variable, the possible outcomes for X are 0,1,2,...100.

    Write it as:
    (0) (probability 0 out of 100 rounds are odd) + (1)(probability 1 out of 100 rounds are odd) ... + (100)(probabiiity 100 out of 100 rounds are odd)

    As @andrewkirk suggests, look up "binomial random variable" if you are unfamiliar with the expressions for these probabilities.
     
  5. Nov 10, 2018 #4
    Thank you both for your answers
     
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