Given two stations, station A transmit in range time [0, T1], and station B transmit in range time [0,T2]. And let's assume that the transmission times for station A and station B are uniformly distributed within their respective time ranges. What is the probability that station A transmit before B while T2 > T1 ?
So as we can see here , the answer is :
1 - $\dfrac{T_1}{2T_2}$
But now I want to calculate the probability that B transmit before A... I thought to calculate it in this way: in half of the options from [0,T1] divide by total of T2 options, namely: $\dfrac{0.5T1}{T2}$ = $\dfrac{T1}{2T2}$. But then we get that:
P(A before B) + P(B before A) + P(B in the same time like A) = 1
1 + P(B in the same time like A) = 1
P(B in the same time like A) = 0
And it's not make sense that this case will be in probability 0..
Can you please help me to understand where is my mistake? And what is the correct way to solve it?
Thank you very much
It is indeed true that
P(B in the same time like A) = 0because you're dealing with continuous distributions, not discrete ones.For a continuous distribution, the probability of a single point is always 0.
To understand this, for a uniform distribution over [0, T1] for instance, the probability that you draw a number in [x, x+dx] is dx/T1. As dx gets closer to 0 (so as you get closer and closer to the probability of getting x exactly), this probability converges to 0.
Another way to understand it is to consider that even if A and B emit "almost at the same time", there always will be some decimal that differs between their two emission times.