Interpretation of network probability problem (transmission delay)

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According to the wiki, the transmission delay is the amount of time required to push all the packet's bits into the wire.

So I understand Probability $P (d_{tr,1} = k ∆_t)$ is that it takes $k$ times to transmit the entire a packet itself by device 1 in $P (d_{tr,1} = k ∆_t)$ chance. (Fail to send a packet $(k-1)$ times and Succeed to send it last one time $\geq k-1 + 1 = k$)

And from problem (a), I intuitively solve this question by inserting $1$ into $k$ because $∆_{t}$ means $k$ is $1$. But I can't understand its solution. I don't get how can it possible to divide this probability into two parts by using inequation such $P (t_{0} + d_{tr,1} ≤ t_{0} + ∆_t)$

I think $d_{tr,1}$ and $k∆_t$ have same meaning 'the transmission delay of the packet'. But this solution makes me confuse. I guess it tries to use cumulative geometrical distribution for solving problem and I cannot step more further unless understand the meaning of two symbols $d_{tr,1}$ and $k∆_t$.

What is the relationship between them? And how I can interpret this equation?

$P(t_{0} + d_{tr,1} ≤ t_{0} + ∆_t ) = P(d_{tr,1} ≤ ∆_t )$

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  • $t_0$ is the time transmission begins
  • $d_{tr,1}$ is the transmission delay time for $1$ packet.
  • $\Delta_t$ is the length of one time slot.
  • $t_0+d_{tr,1}$ is the time that $1$ packet arrives; time of transmission plus transmission delay.
  • $t_0+\Delta_t$ is the latest time we desire that packet to arrive; time of transmission plus one time slot.
  • $\{t_0+d_{tr,1}\leqslant t_0+\Delta_{1}\}$ is the event that $1$ packet is delayed no later than one time slot after transmission begins. We seek the probability of this event.
  • $\{d_{tr,1}\leqslant \Delta_{1}\}$ is the event that $1$ packet is delayed no longer than one timeslot. We can calculate the probability of this event.
    • These should be the same event as transmission delay is presumably independent of time of transmission.
  • $\therefore~P(t_{0} + d_{tr,1} ≤ t_{0} + ∆_t ) ~=~ P(d_{tr,1} ≤ ∆_t )$

That is all.