Probability that minimum of two numbers is less than 4 Suppose I have to choose two numbers from set $$S=\{1,2,3,4,5,6 \}$$ without a replacement , then what is the probability that minimum of two is less than $4$?
I made two groups for this problem $A= \{1,2,3 \}$ and $B=\{4,5,6 \}$
.There are two possibilities , either both are from $A$ or one from $A$ and one from $B$ to satisfy our requirement. Hence
$$P(E)=\frac{3}{6}.\frac{2}{5} + \frac{3}{6}.\frac{3}{5}$$
but answer is incorrect. Help?

It is easy as P.I.E.. Use the Principle of Inclusion and Exclusion.
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = \mathsf P(X < 4)+\mathsf P(Y < 4)- \mathsf P(X< 4 \cap Y< 4)
\\ &= \frac 3 6 + \frac 3 6 - \frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5 
\end{align}$$
Or you can use the Law of Complements
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = 1- \mathsf P(X \geq 4 \cap Y\geq 4)
\\ &= 1 - \frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5 
\end{align}$$
Or by your method (the Law of Total Probability), but remembering that there are two ways for one pick to be in A and the other in B.
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = \mathsf P(X < 4 \cap Y\geq 4)+\mathsf P(X \geq 4 \cap Y< 4) + \mathsf P(X< 4\cap Y< 4)
\\ & = \frac 3 6\cdot\frac 3 5 +\frac 3 6\cdot\frac 3 5 +\frac 3 6\cdot\frac 2 5  
\\ & = \frac 4 5 
\end{align}$$
