Finding the probability function

Question

Jug contains $8$ balls, $3$ are red and $5$ are blue.pulling out balls with no return intill the first blue ball comes out

$\underline{\color{red}\bullet\times3 \,\,\,\,\,\color{blue}{\bullet}\times5}$

Find the probability function of the number of balls that we need to pull out from the jug

My try:

Let $X$ be the number of balls that we need to pull out from the jug

$$ \begin{array}{c|lcr} X&1&2&3&4 \\ \hline \text{P}_\text{X}(x) & 5/8 & 5/7 &5/6 &5/5 \\ \end{array} $$

But $\Sigma P_{X}(x)>1$

Answer 1

To obtain the first blue ball on draw $x$ you need to select $x-1$ of the three red balls, and then one of the five blue balls; out of all the ways to select $x$ of the eight balls. $$\mathsf P_X(x) = \dfrac{\dbinom{3}{x-1}\dbinom{5}{1}}{\dbinom{8}{x-1}\dbinom{9-x}{1}}\cdot\;\mathbf 1_{x\in\{0,1,2,3,4\}}$$

Answer 2

Multiply each $(1 - P_x)$ and subtract from 1.

Answer 3

I think the first draw is clear. For the number of draws to equal 2, then what must happen is that you got a RED then a BLUE $$\frac{3}{8}\cdot\frac{5}{7}=\frac{15}{56}.$$ The same logic is applied to the 3rd and 4th draw. $$ \begin{array}{c|lcr} x&1&2&3&4 \\ \hline \text{P}(X=x) & 35/56 & 15/56 & 5/56 & 1/56 \\ \end{array} $$

Answer 4

$P(X=1)=\frac{5}{8}, P(X=2)=\frac{3}{8}\frac{5}{7},P(X=3)=\frac{3}{8}\frac{2}{7}\frac{5}{6}, P(X=4)=\frac{3}{8}\frac{2}{7}\frac{1}{6}\frac{5}{5}$.

Namely, for each $k$ you have to consider all the previous tries before the first "blue" ball.