"A man is playing versus a machine in the following way: The machine chooses 2 numbers randomly from the set of numbers 1,2,3,4,5, where a number can be chosen twice (with replacement). If the multiplication of the 2 chosen numbers is even, the man gets 5 dollars Calculate the expected value (mean) and variance of the profit after 100 games, if for every game he pays 2 dollars to play."

What I did to start with, is to calculate the probability of having an even multiplication and I got p=16/25. Now I know I can calculate the profit for a single game, get a probability function with 2 values, and find E(X), and multiply it by 100. I am not sure if I can do the same with the variance, and I think I might be able to do it with Binomial distribution, but not sure how. Maybe a linear transformation is also possible?

  • $\begingroup$ I have added the [self-study] tag. These questions are fine, but please read the wiki to understand the process. $\endgroup$ Commented Jan 6, 2015 at 7:30
  • $\begingroup$ If X is the profit in a single game, then it's possible values are: 3 and -2. I know the matching probabilities, which are 16/25 and 9/25 respectively. Therefore I can find E(X). The expected value for 100 games is simply 100*E(X). All I ask is how to do the variance, is it simply 100*V(X) ? Or 100 squared by V(X) ? $\endgroup$
    – user64983
    Commented Jan 6, 2015 at 7:53
  • $\begingroup$ en.wikipedia.org/wiki/Variance#Basic_properties $\endgroup$
    – Glen_b
    Commented Jan 6, 2015 at 14:32

1 Answer 1


So this doesn't remain unanswered:

In simplest terms, this merely relies on basic properties of variance, specifically, for independent random variables $X_1,X_2,...,X_n$,

$\text{Var}(\sum_{i=1}^n X_i)=\sum_{i=1}^n \text{Var}(X_i)\,$.


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