Sick and healthy persons

Question

We pick 2 random people. The chance of a person being sick is 0.5 and the chance of a person being healthy is 0.5.

Our random variable is the number of sick people.

The expected value is 1 correct?

Answer 1

Yes, it is 1 as you mention. Let's see why in a more general framework that one should follow to tackle such problems. First, recognize what type of distribution your random variable follows, in this case, it is a Binomial distribution with $n=2$ and $p=0.5$. You are interested in the mean or the expected value of this random variable, this is given as $n*p$ which as you can see in the Wiki article. In our example, it would just be $n*p$ so $2*0.5=1$.

Answer 2

Yes, you´re right. You can calculate the expected value straight forward.

$$\mathbb E(X) =\sum_{k=0}^2 k\cdot {2 \choose k} \cdot 0.5^k\cdot 0.5^{2-k}=\sum_{k=0}^2 k\cdot {2 \choose k} \cdot 0.5^2$$

$$=0\cdot {2 \choose 0} \cdot 0.5^2+1\cdot {2 \choose 1} \cdot 0.5^2+2\cdot {2 \choose 2} \cdot 0.5^2=0.5^2\cdot \left(0+2+2\right)=0.25\cdot 4=1$$