# Understanding the Probability that A realizes before B in a mission whose duration is T unit

$$A$$: random variable to denote time to failure for equipment E_A

$$B$$: random variable to denote time to failure for equipment E_B

$$A$$ and $$B$$ are independent.

$$P(A◁B)(T)$$ : Probability that A realizes before B in a mission whose duration is T unit.

example scenario:

$$A$$ and $$B$$ both exhibit exponential distribution.

E_A failure rate: $$\lambda_A=5E-3$$ /hours

E_B failure rate: $$\lambda_B=9.4E-3$$ /hours

duration: $$100$$ hours

what is $$P(A◁B)(T)$$ ?

I tried: (found equation below in a paper. I can understand the equation.)

$$P(A◁B)(T)=\int_0^T \biggl( pdf_A(t)*(1-CDF_B(t)) \biggr) dt$$

my questions:

1. integral result is 0.2650 if $$\int$$ time range is $$0$$ to $$100$$
2. integral result is 0.3472 if $$\int$$ time range is $$0$$ to $$\infty$$
3. Which one is correct, $$0.2650$$ or $$0.3472$$? Why?

my confusion to ask question 3 is why result depends on time? I feel that only the values of $$\lambda$$s are significant here since exponential rate is constant and exponential distr. is memoryless.

• sure. 1st mission (0 to100)=26.5% , 2nd mission: 6.16% , 3rd mission 0.0616 , 4th mission 0.0146 and so on. Sum of these probabilities is 34.72%. In my real life related question I don't have any $T_0$ or $T_1$. I just have duration Commented Feb 18, 2022 at 7:53
$$\begin{matrix} 1. & & & & & A < B \ \text{ and } A < T \\[6pt] 2. & & & & & A < B. \\[6pt] \end{matrix}$$