Given an urn with $40$ balls: $10$ white,$10$ black,$10$ red, $10$ green. We extract $2$ balls (simultaneously). What is the probability that at least one of them to be a white ball?
 A: Look at the event $X$= 
you dont pick a white ball.
Numbers of combinations to pick two balls : $40 \choose 2 $
Number of combinations to pick two balls that are not white : $30 \choose 2 $
Therefore the probability to pick two non-white balls is $$P(X)=\frac{30 \choose 2 }{40 \choose 2 }$$
We know that $X^c$= You pick at least one white ball, thus
$$P(X^c)=1-\frac{30 \choose 2 }{40 \choose 2 }$$
A: Solve by counting:


*

*Direct: "At least one white" = ("Two white") or ("One white and one non-white"): $$
\frac{\binom{10}{2}}{\binom{40}{2}}+\frac{\binom{10}{1}\binom{30}{1}}{\binom{40}{2}}=\frac{23}{52}
$$ Where $P(X)=P(X_1\text{ or }X_2)=P(X_1)+P(X_2)$ since $X_1,X_2$ are disjoint events.

*Complement: "Pick $x\ge 1$ white" is opposite of "Pick $x\lt 1$ white (that is, pick $0$ whites)": $$
1-\frac{30 \choose 2 }{40 \choose 2 }=\frac{23}{52}
$$ Where $P(X)=P(\text{not }X_0)=1-P(X_0)$ due to $X_0$ being a complement of $X$.
Solve by probability tree:


*

*Chances of first ball being white is $\frac{10}{40}$, and of not being white is $\frac{30}{40}$.

*If first is white, we are done (don't care about the second pick as we already have at least one white in this scenario). If first is not white (And don't forget we have $\frac{30}{40}$ chance of finding ourselves in this scenario), we now have $\frac{10}{39}$ of second ball being white (first pick took out one non-white ball). Summing up both scenarios: $$
\frac{10}{40}+\frac{30}{40}\cdot\frac{10}{39}=\frac{23}{52} $$
