Find the expected value and the variance of the weight of a person In room number one there are three women that weigh $61,57,59$ kg In room number two there are three men that weigh $68,72,73$ kg choosing one room randomly and picking a person, > $\color{blue}{A)}$ find the expected value of the wight of a person and the variance > > $\color{blue}{B)}$ find the expected value of the wight of a person and the variance if we know that the person is from room number one My try: $$\color{blue}{A)}\mathbb{E}(X)=\frac{1}{6}(61+57+59+68+72+73)=\boxed{65}$$ $$\mathbb{V}(X)=\frac{1}{6}[(61-65)^2+(57-65)^2+(59-65)^2+(68-65)^2+(72-65)^2+(73-65)^2]$$ $$\color{blue}{B)}\mathbb{E}(X)=\frac{1}{3}(61+59+57)=\boxed{29.5}$$ $$\mathbb{V}(X)=\frac{1}{3}[(61-65)^2+(57-65)^2+(59-65)^2]$$ > My attempt is correct?

First of all, you are using the same name $X$ for both A & B. I suggest $E(X)$ and $V(X)$ for A and $E(X|R=1)$ and $V(X|R=1)$ for B, where $R$ is the room.

Second, I think you are computing the expectation in B incorrectly, you wrote division by 3 but actually divided by 6. The actual expectation is $E(X|R=1)=59$.

Accordingly, you need to subtract 59, not 65 from each value when you are computing $V(X|R=1)$.