Let $X ∼ Bin(n,\frac12)$ and $Y ∼ Bin(n + 1,\frac12),$ independently. Let $V = \min(X, Y )$ and $W = \max(X, Y )$. Find $E(V ) + E(W ).$
What I have tried so far: Let $q$ be the probability that $x$ is smaller, then $y$ will have the probability of $1-q.$ Then $E(\min(X,Y)) = xq + y(1-q).$ Similarly, if $y$ is the maximum with probability $q$, then $x$ will have probability $1-q$ so that $E(\max(X,Y)) =x(1-q) + yq.$ That is $$E(\min(X,Y)) + E(\max(X,Y)) = x + y$$ Then $E(x) = np$ and $E(y) = (n+1)p.$
$np + (n+1)p = (2n + 1)p$ We know that $p= 1/2,$ hence $(2n + 1)p = n + 1/2.$ That is what I have but I am not sure if that is the answer?
You're correct but we can make it easier: $$ \text E[V] +\text E[W] = \text E[V+W] = \text E[X + Y] = \text E[X] + \text E[Y] = n + \frac 12. $$ $X + Y \sim \text{Bin}(2n+1, 1/2)$ by independence and the common success probability but we don't even need this due to the linearity of expectation.