A clown moves forward in jumps of 1 unit to the right or one unit to the left. He jumps left with P=p and right with $P=1-p$. With independence between jumps. Let $C$ be the position of the clown after n jumps. What is $E[C]$ and $V[C]$.
I basically said the clown will be np jumps to the left and $n(1-p)$ to the right. So $E[C]=np-n(1-p)=2np-n$. Where a positive expectation means the clown will be positioned to the left. Does this seem correct? And would variance be $np(1-p)$?
Let $X_i=-1$ if the $i$-th jump is to the left, and $X_i=1$ if the $i$-th jump is to the right. Then $C=X_1+\cdots+X_n$. The expectation of $C$ is the sum of the expectations of the $X_i$. The expectation of any $X_i$ is $(p)(-1)+(1-p)(1)$.
For the variance, by independence the variance of $C$ is the sum of the variances of the $X_i$. I am sure you can find the variance of $X_i$.
Another way: Consider $Y=\frac{1-C}{2}$. This has binomial disribution, probability of "success" equal to $p$. So by standard facts the mean of $Y$ is $np$ and the variance of $Y$ is $np(1-p)$.
Note that $C=1-2Y$. Now we can read off the mean and variance of $C$ from the corresponding facts about $Y$.