# Expected Value: Use Indicator variables and random variables

Question:

Every time a customer orders a drink, the waiter serves the wrong drink with probability $$\frac{1}{12}$$, independently of other orders.

You order $$7$$ ciders, one cider at a time. Let $$(D_1,D_2,...,D_7)$$ be the sequence of drinks that the waiter serves. De fine the following random variable $$X$$:

$$X$$ = the number of indices i such that $$D_i$$ is a cider and $$D_{i+1}$$ is not a cider.

What is the expected value $$E(X)$$ of $$X$$

Answer: 0.45833333

Attempt: I start off labelling my indicator variable: $$X = \left\{\begin{array}{rc} 1,&\text{the number of indices i such that D_i is a cider and D_{i+1} is not a cider}{} \\ 0,&\text{any other cases}{}\end{array}\right.$$

I need to find $$P(X=1)$$ but I'm not sure how to go about it. Will I be served 4 ciders and 3 non-cider drinks according to the condition? How do I incorporate the probability of the waiter getting the drink wrong?

I am struggling to proceed with these questions after defining the indicator variables. Any step by step guide would be appreciated.

## 1 Answer

Let $$X_i$$ be indicator variable for $$i$$-th event, so $$X=X_1+...+X_7$$. Then , for $$i\leq 6$$ we have $$E(X_i)=P(X_i=1) = {11\over 12}\cdot {1\over 12}$$ and $$E(X_7)= P(X_7 = 1)= {11\over 12}\cdot 0 = 0$$

so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6\cdot {1\over 12}\cdot {11\over 12} = {11\over 24}$$