Exponential random variables with average number of beeps A device randomly beeps, with an average of 2 beeps per second.

If we assume that the time $T$ between any two beeps is an exponential random variable, then what is the probability that no beep is emitted in the next 5 seconds? Now, if we let $Y_i$ be exponential random variables modeling the time between the $i$th and the $i$ + 1st beeps, what is the probability that there are exactly 3 beeps in the next 4 seconds? I know we can parametrize this with a parameter $\lambda$ which is 2 beeps per second, but I am unsure of how to proceed in each of these situations.
Preliminary
Gamma distribution to the rescue! If a R.V. $T\sim Gamma(a,\mu)$, its pdf is...
$$
P[T = t] =
\begin{cases}
\frac{t^{a-1}e^{-\frac{t}{\mu}}}{\mu^a\Gamma(a)},  &amp; t \gt 0 \\
0, &amp; t \le 0
\end{cases}
$$
And its cdf is...
$$
P[T &lt; t] =
\begin{cases}
\frac{\gamma(a,\frac{t}{\mu})}{\Gamma(a)},  &amp; t \gt 0 \\
0, &amp; t \le 0
\end{cases}
$$
Where $\mu$ is the average per trial and $a$ is the # inter-arrival times.
Answer 1
For your problem, $a = 1$ (The gamma reduces to the exponential distribution) and $\mu = 0.5$ [s/beep]...
$$
P[T&gt;5] = 1-P[T&lt;5] = 1 - \frac{\gamma(1,2\cdot 5)}{\Gamma(1)} = 1 - (1 - e^{-10})
\\\implies P[T&gt;5] =e^{-10}
$$
Answer 2
Let $Y = Y_1 + Y_2 + Y_3$ so $a = 3$ this time...
$$
P[Y=4] = \frac{4^{3 - 1}e^{-2\cdot 4}}{0.5^3\Gamma(3)}
\\\implies P[Y=4] = 64e^{-8}
$$
Preliminary
Gamma distribution to the rescue! If a R.V. $T\sim Gamma(a,\mu)$, its pdf is...
$$
P[T = t] =
\begin{cases}
\frac{t^{a-1}e^{-\frac{t}{\mu}}}{\mu^a\Gamma(a)},  &amp; t \gt 0 \\
0, &amp; t \le 0
\end{cases}
$$
And its cdf is...
$$
P[T &lt; t] =
\begin{cases}
\frac{\gamma(a,\frac{t}{\mu})}{\Gamma(a)},  &amp; t \gt 0 \\
0, &amp; t \le 0
\end{cases}
$$
Where $\mu$ is the average per trial and $a$ is the # inter-arrival times.
Answer 1
For your problem, $a = 1$ (The gamma reduces to the exponential distribution) and $\mu = 0.5$ [s/beep]...
$$
P[T&gt;5] = 1-P[T&lt;5] = 1 - \frac{\gamma(1,2\cdot 5)}{\Gamma(1)} = 1 - (1 - e^{-10})
\\\implies P[T&gt;5] =e^{-10}
$$
Answer 2
Let $Y = Y_1 + Y_2 + Y_3$ so $a = 3$ this time...
$$
P[Y=4] = \frac{4^{3 - 1}e^{-2\cdot 4}}{0.5^3\Gamma(3)}
\\\implies P[Y=4] = 64e^{-8}
$$