Probability of three successes before three failures with additional conditions There is a random variable $X$ that takes integer values from 1 to 20. 
An event $A = \{10 \leq X \leq 19\}$ counts as a success.
An event $B = \{2 \leq X \leq 9\}$ counts as a failure.
An event $C = \{X = 20\}$ counts as three successes.
An event $D = \{X = 1\}$ counts as two failures.
What is the probability that three successes happen before three failures?
My solution
The probability that there are at least three failures equals $p=P(D)^2+P(D)P(B)+P(B)^3=\frac{1}{20}\frac{1}{20}+\frac{1}{20}\frac{8}{20}+\left(\frac{8}{20}\right)^3=\frac{173}{2000}$
I then count the probability of zero, one and two successes.
$P\{\text{there are no successes}\}=p=\frac{173}{2000}$
$P\{\text{there is only one success}\}=p*P\{A\}=\frac{173}{2000}\frac{1}{2}$
$P\{\text{there are only two successes}\}=p*P\{A\}^2=\frac{173}{2000}\frac{1}{4}$
So, my probability equals 
$P(Q)=1 - \frac{173}{2000}\frac{7}{4}=\frac{6769}{8000}\approx84.86\%$
Am I right?

Probabilities:


*

*Single Success: $1/2$

*Triple Success: $1/20$

*Single Failure: $2/5$

*Double Failure: $1/20$


To get three successes before three failures you must obtain one of the following disjoint sequences:
$$\boxed{\begin{array}{l:l}
\text{A triple success.}&\tfrac 1{20}\\ \hdashline
\text{ Three single successes }&{(\tfrac 12)}^3\\ \hdashline
\text{A single failure and two successes, then a single success.} &\tbinom{3}{1}(\tfrac{2}{5}){(\tfrac{1}{2})}^3\\ \hdashline
\text{ A single failure, then a triple success.}& \\ \hdashline
\text{ } &  \\ \hdashline
\text{ } & \\ \hdashline
\text{ }& \\ \hdashline
\text{ }& \end{array}}$$
Complete the table, sum and simplify.
