Using addition theorem in conditional probability Two balls are drawn from an urn containing 2 white,3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red?
 A: Your approach is valid, but there is a quicker way -- especially if we decide to increase the number of balls we draw.
Recall that $P(\text{at least one red ball}) = 1-P(\text{no red balls})$.
The probability that we don't draw a red ball the first time is $6/9$, since there are $6$ non-red balls, and $3$ red balls.
Now there are $8$ balls remaining: $5$ non-red balls and $3$ red balls. The probability that I don't draw a red ball the second time is $5/8$.
So $$P(\text{at least one red ball})=1-\left(\frac{6}{9}\right)\left(\frac{5}{8}\right).$$
A: If we let $R_n$ mean the event of the $n$-th ball being red, and $N_n$ mean the event of the $n$-th ball being not red (either black or white), then you want
$$\begin{align}
\mathsf P(R_1\cup R_2)
 & = \mathsf P(R_1\cap R_2) + \mathsf P(R_1\cap N_2) + \mathsf P(N_1\cap R_2)
\\[1ex] & = \mathsf P(R_1) + \mathsf P(N_1\cap R_2)
\\[1ex] & = \mathsf P(R_1) + \mathsf P(N_1)\,\mathsf P(R_2\mid N_1)
\end{align}$$
The event of at least one ball being red is either that the first ball is red, or that the first is not red and the second ball is.
An equivalent expression is obtained by the law of complements
$$\begin{align}
\mathsf P(R_1\cup R_2)
 & = 1 - \mathsf P(N_1\cap N_2)
\\[1ex] & = 1 - \mathsf P(N_1)\,\mathsf P(N_2\mid N_1)
\end{align}$$
Now, do you know how to calculate these probabilities?  
A: Use the Hypergeometric distribution. If a random variable $X$ has Hypergeometric distribution: $X \stackrel{d}{=}Hg(n,R,N)$. Where $N$ is the size of the population, $n$ is the sample you take, and $R$ is the number of individuals from the population with the characteristic you are looking for. Then we have:
$$P(X=x)=\frac{\binom{R}{x}\binom{N-R}{n-x}}{\binom{N}{n}}.$$
You need to find $P(X\geq1)=1-P(X=0)$, $X \stackrel{d}{=}Hg(n=2,R=3,N=9).$
