An urn contains 15 Balls (5 white, 10 Black). Let's say we pick them one after the other without returning them. How many white balls are expected to have been drawn after 7 turns?
I can calculate it by hand with a tree model but is there a formula for this?
On
An easier method: approach via indicator random variables.
Let $\chi_n = \begin{cases} 1&\text{if the}~n\text{'th ball is white}\\0&\text{otherwise}\end{cases}$
Let $X=\sum\limits_{n=1}^7\chi_n$. That is to say, $X$ counts the number of white balls in the first seven draws.
We wish to calculate $E[X]$
By the linearity of expectation:
$$E[X]=E\left[\sum\limits_{n=1}^7\chi_n\right]=\sum\limits_{n=1}^7E[\chi_n]$$
Now, recognize that $E[\chi_n]=E[\chi_1]$ for every $n\in\{1,2,\dots,7\}$. The probability that the first ball is white is $\frac{5}{15}$, hence $E[\chi_1]=\frac{5}{15}\cdot 1 + \frac{10}{15}\cdot 0 = \frac{5}{15}$
Why was $E[\chi_n]=E[\chi_1]$ for all $n$? By temporarily assigning labels to each ball, we see that there are $15!$ ways to arrange the balls in a line. The first seven balls in the line will be those seven which we will be pulling. Each of these $15!$ arrangements are equally likely to occur. In exactly $\frac{5}{15}$ of these arrangements (that is to say, in exactly $5\cdot 14!$ arrangements) the first ball is white. Similarly, in exactly $\frac{5}{15}$ of these arrangements the second ball is white, etc... In general, in precisely $\frac{5}{15}$ of the arrangements the $n$ 'th ball is white. Hence the conclusion.
We have then:
$$E[X]=\sum\limits_{n=1}^7E[\chi_n]=\sum\limits_{n=1}^7E[\chi_1]=7E[\chi_1]=7\cdot\frac{5}{15}=\frac{7}{3}\approx 2.3333\dots$$
Hint: the linearity of expectations show us that
$$\Bbb E[W]=\sum_i\Bbb E[X_i]$$
where $W$ is the random variable that counts the total number of white balls, and each $X_i$ is the random variable that take value $1$ if the ball is white or zero if the ball is black for the $i$-nth pick.
The $X_i$ random variables are not independent, by example, if the first ball is white ($X_1=1$) the expectation of $X_2$ is different that if the first ball is black.
Then we use the total law of probability i.e.
$$\Pr[X_i=1]=\sum\Pr[X_i=1|X_{i-1}=a_{i-1},X_{i-2}=a_{i-2},...,X_1=a_1]\cdot\Pr[X_{i-1}=a_{i-1},X_{i-2}=a_{i-2},...,X_1=a_1]$$
where the sum ranges all combinations of $a_i$, where every $a_i\in\{0,1\}$, i.e. every ball can be white or black. This is just a bit of combinatorics. Check this answer that is similar to understand how to continue.