An urn contains four white balls and six black. Three balls are drawn with replacement. Let $x$ be the number of white balls. Calcaulate $E (x)$, $VAR(x)$ and $\sigma x$.
I don't know how to calculate $E(x) =\sum\limits_{i=1}^{n} X_i P(X_i)$.
An urn contains four white balls and six black. Three balls are drawn with replacement. Let $x$ be the number of white balls. Calcaulate $E (x)$, $VAR(x)$ and $\sigma x$.
I don't know how to calculate $E(x) =\sum\limits_{i=1}^{n} X_i P(X_i)$.
Let me give you some hints: let $X$ be a number of white balls
Use combinatorics to find $p_0 = P(X = 0)$, $p_1 = P(X = 1)$, $p_2 = P(X = 2)$, $p_3 = P(X = 3)$. For example, the probability $p_0$ that there are no white balls at all is $b^3$ where $b = \frac{6}{10}$ is a probability to draw a black ball.
For $E(X)$ you have $E(X) = \sum\limits_{k=0}^3 k\cdot p_k = p_1+2p_2+3p_3$.
Because you use replacement the probability of white will be 4/10 =2/5=0.40 each time you draw. Each draw is independent of previous ones so for a given sequnce of white and black you can get the probability by multiplication. So for example the sequence W B W has probability that is the following product: (2/5)x(3/5)x(2/5). Since this os the set up the number of w balls drwn in three draws has a binomial distribution with p=2/5 and n=3. With all those hints you should be able to solve this.