Probability of an event occurs at least three times before another event occurs cars arrives according to a Poisson process with rate=2 per hour and trucks arrives according to a Poisson process with rate=1 per hour. They are independent. 
What is the probability that  at least 3 cars arrive before a truck arrives? 
My thoughts: 
Interarrival of cars A ~ Exp(2 per hour), Interarrival of trucks B ~ Exp(1 per hour). 
Probability that  at least 3 cars arrive before a truck arrives
$= 1- Pr(B<A) - Pr(A<B)Pr(B<A) - Pr(A<B)Pr(A<B)Pr(B<A)
\\= 1 - (\frac{1}{3})-(\frac{2}{3}\cdot\frac{1}{3})-(\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3})\\=\frac{8}{27}.$ 
Is this correct?
 A: Let $M_t$ be the Poisson process which counts the arrival of trucks. Then by the given condition we have $(M_t)_{t>0} \sim PP(1).$ Let $X_1,X_2, X_3, \cdots$ be the time gaps between arrival of cars. Then $X_n \sim \text{iid} \exp (2).$ So the required probability is $P(M_{X_1+X_2+X_3} < 1).$
Now $$\begin{align} 
P(M_{X_1+X_2+X_3} < 1) & = P(M_{X_1+X_2+X_3} = 0). \\ & = \int_{0}^{\infty} P(M_{X_1+X_2+X_3} = 0 \mid X_1+X_2+X_3 = t) f_{X_1+X_2+X_3} (t)\ \text{dt}. \\ & =  \int_{0}^{\infty} P(M_t = 0) f_{X_1+X_2+X_3} (t)\ \text{dt}. \end{align}$$
Now $M_t \sim \text {Poisson}\ (t)$ and $X_i$'s are iid with exponential$(2)$ it follows that $X_1+X_2+X_3 \sim \text {Gamma} (3,2).$
So $$\begin{align} P(M_{X_1+X_2+X_3} < 1) & = 4 \int_{0}^{\infty} t^2e^{-3t}\ \text {dt}. \end{align}$$
Using gamma function we find that $$\int_{0}^{\infty} t^2 e^{-3t}\ \text{dt} = \frac {\Gamma(3)} {27} = \frac {2} {27}.$$
So the required probability is $4 \times \frac {2} {27} = \frac {8} {27},$ as you have obtained.
A: Note that the waiting times for the poisson distribution is an exponential distribution with mean $\lambda$. Letting $X$ denote the time it takes for the first car to arrive and $Y$ the time it takes for a truck to arrive, we have 
$$\mathsf P(X\lt Y)=\frac{\lambda_X}{\lambda_X+\lambda_Y}=\frac{2}{3}$$
Since the exponential distribution is memoryless, the probability that we observe $3$ or more cars before a truck arrives is simply $$\left(\frac{2}{3}\right)^3=\frac{8}{27}$$
A simulation using R statistical software agrees with this result
> x1<-rexp(10^7,2)
> x2<-rexp(10^7,2)
> x3<-rexp(10^7,2)
> y<-rexp(10^7,1)
> mean(x1+x2+x3<y)

[1] 0.2963617

> 8/27

[1] 0.2962963

