Random Variable - Mean and Variance

[EEristavi]

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 can't 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 σ_s=√(100 * 0.5 * 0.5) = 5. However, I can't understand how we get this

 

[andrewkirk]

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.

 

[Stephen Tashi]

I understand that the probability - Pr = 0.5.

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.

I can't figure out how to write it appropriately (use formula E[x] = ∑ x * Pr)

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.

 

[EEristavi]

Thank you both for your answers