What is the probability of having one red ball given that at most two of the selected balls are red? A box contains 4 red and 5 white balls. Three balls are selected randomly. What is the probability of having one red ball given that at most two of the selected balls are red?
I create a pdf table as following
$X|\;\,0\;\, |\;\,1\;\,|\;\, 2 \;\, | \;\, 3\;\, |$
$P|\,\frac5{42}\;|\,\frac{20}{42}\;|\,\frac{15}{42}|\,\frac{2}{42}\;|$
And I have an answer but I can't continue
$P(1R\;|<3R)= (...)$
 A: You are on the right track, but your reasoning needs refinement.
What you calculated are the probabilities of drawing $X$ number of red balls, where $X$ ranges from $0$ to $3$.  What is worth noting from your table is that $\Pr[X = 3] = \frac{1}{21}$; that is to say, the probability that all three balls drawn are red is $$\frac{\binom{4}{3}\binom{5}{0}}{\binom{9}{3}} = \frac{1}{21};$$ this is because there are $\binom{4}{3} = 4$ ways to choose $3$ red balls from $4$ in the box; $\binom{5}{0} = 1$ way to choose $0$ white balls from the $5$ in the box; and $\binom{9}{3} = 84$ ways to choose any three balls from the box.
The reason why I say your reasoning needs refinement is that we can see that in fact, there are $4$ such elementary outcomes out of the $84$ total possible.  The other $84 - 4 = 80$ outcomes correspond to the condition that there are at most $2$ red balls drawn; i.e., "at most two red balls drawn" is equivalent to saying $X \le 2$.  And of these, how many correspond to having exactly one red ball?  Well, you already have it in your table, if you instead count outcomes instead of probabilities:
$$\Pr[X = 1] = \frac{10}{21} = \frac{40}{84}$$ implies there are $40$ outcomes in which exactly one ball is drawn.  And out of the $80$ that correspond to at most two red balls drawn, the answer is simply $$\Pr[X = 1 \mid X \le 2] = \frac{40}{80} = \frac{1}{2}.$$
In fact, this is precisely what evaluating the conditional probability means:  $$\Pr[X = 1 \mid X \le 2] = \frac{\Pr[(X = 1) \cap (X \le 2)]}{\Pr[X \le 2]} = \frac{\Pr[X = 1]}{1 - \Pr[X = 3]}.$$  We reason that the event $X = 1$ is a proper subset of the event $X \le 2$, thus their intersection is simply $X = 1$; and the event $X \le 2$ is the complement of the event $X = 3$.
All we did when we counted outcomes was make it more relatable to the events concerned, rather than working with the probabilities.  For if your table had looked like this:
$$\begin{array}{c|c|c|c|c}
X & 0 & 1 & 2 & 3 \\ \hline
\text{Number of outcomes} & 10 & 40 & 30 & 4 
\end{array}$$
I imagine it might have been easier to understand how to get the desired conditional probability.

To further help your understanding, how would you compute the probability that at least $2$ white balls were drawn, given that the number of red balls drawn was an odd number?
For another exercise, suppose the box also contains $6$ green balls, and you draw $5$ balls at random without replacement.  What is the probability that $1$ ball drawn is red, $2$ are white, and $2$ are green?
Finally, given that exactly $2$ balls drawn are green, what is the probability that you drew at most $1$ red ball?
