What is the probability that the phone will ring (i.e., the first call will arrive) within the next $30$ minutes (i.e. within the next $1/2$ an hour)? Dan, Dominic, and Doug
are waiting together in the living room for their girlfriends, Sally, Shellie, and
Susanne, to call. Their waiting times (in hours) are independent Exponential
random variables, with parameters $2.1$, $3.7$, and $5.5$, respectively. What is the
probability that the phone will ring (i.e., the first call will arrive) within the
next $30$ minutes (i.e., within the next $1/2$ an hour)?
so $\lambda_1 = 1/126$  for Dan
$\lambda_2 = 1/222$ for Dominic
$\lambda_3 = 1/330$ for Doug
Let X represent Dan's waiting time. Y represent Dominic's waiting time. Z represent Doug's waiting time.
$P(X\leq30) = 1 -e^{-30/126}$
$P(Y\leq30) = 1 -e^{-30/222}$
$P(Z\leq30) = 1 -e^{-30/330}$
so $P(min (X,Y,Z) \leq 30) = ???$
I don't know what to do in this last step. The final answer is  $0.9965$ But how did they get that answer?
 A: Let's consider two independent exponential random variables, each with a different rate parameter, say $X \sim \operatorname{Exponential}(\lambda)$, $Y \sim \operatorname{Exponential}(\theta)$, with $$F_X(x) = 1 - e^{-\lambda x}, \quad F_Y(y) = 1 - e^{-\theta y}.$$  We construct a new random variable $Z = \min(X,Y)$.  How is $Z$ distributed?  We note $$\Pr[Z > z] = \Pr[\min(X,Y) > z] = \Pr[(X > z) \cap (Y > z)] \overset{\text{ind}}{=} \Pr[X > z]\Pr[Y > z] = e^{-\lambda z} e^{-\theta z} = e^{-(\lambda + \theta)z}.$$  Hence $$F_Z(z) = \Pr[Z \le z] = 1 - \Pr[Z > z] = 1 - e^{-(\lambda + \theta)z},$$ meaning $Z = \min(X,Y)$ is exponential with rate $\lambda + \theta$, the sum of the rates of $X$ and $Y$.  So if we have three independent exponential random variables, the minimum of these waiting times will also be exponential with rate equal to the sum of the individual rates.
Applying this to the question, we then have the probability that the first call to arrive being within the first $30$ minutes is simply $$\Pr[W = \min(X,Y,Z) \le 30] = 1 - e^{-(\lambda_1 + \lambda_2 + \lambda_3)30} = 1-e^{-0.464139} \approx 0.371324.$$  I do not get the claimed answer; it's too large.
A: Derivation of minimum:
$V = \min(X,Y,Z)$  Use the CDF method to find the distribution of $V.$
$$F_V(v) = P(V \le v) =  1 - P(V > v) \\
= 1 - P(X>v, Y>v, Z>v)\\ = 1 - P(X>v)P(Y>v)P(V>v) \\
= 1 - e^{-\lambda_1v}e^{-\lambda_2v}e^{-\lambda_3v}
= 1 - e^{-(\lambda_+\lambda_2+\lambda_3)v},$$
for $v > 0.$ We recognize this as the CDF of an
exponential distribution with rate $\lambda = \lambda_1+\lambda_2+\lambda_3.$
$P(V \le 30) = 1 - e^{-30\lambda} = 0.3713.$  Computation in R:
lam.1  = 1/126;  lam.2 = 1/222;  lam.3 = 1/330
lam = lam.1 + lam.2 + lam.3;  lam
[1] 0.01547132
1 - exp(-30*lam)
[1] 0.3713241
pexp(30, lam)  3 'pexp' is exponential CDF
[1] 0.3713241

Simulation:
set.seed(2021)
lam.1  = 1/126;  lam.2 = 1/222;  lam.3 = 1/330
# rates per min
x = rexp(10^6, lam.1)
y = rexp(10^6, lam.2)
z = rexp(10^6, lam.3)
v = pmin(x,y,z)
mean( v <= 30)
1 0.629108
2*sd(v <= 30)/1000
1 0.0009660877
mean(v <= 30)
[1] 0.370892
2*sd(v <= 30)/1000
[1] 0.0009660877

So the probability of a call in the next half hour
is $P(V \le 30) \approx 0.371 \pm 0.001.$
hdr = 
 "Histogram of Simulated Dist'n of V 
with Density of EXP(rate=0.0155)"
hist(v, prob=T, br=50, col="skyblue2", main=hdr)
 curve(dexp(x, 0.0155), add=T, col="red", lwd=2)
 abline(v = 30, col="darkgreen", lwd=2, lty="dotted")


A: Hint : you need to find $P(X\leq30  \cup Y\leq30  \cup Z\leq30 )$
A: A much simpler approach.  The desired result is $1-Prob((X\ge30) \cap (Y\ge 30)\cap (Z\ge 30))=$$1-e^{-(30/126+30/222+30/330)}=.3713$
The final result uses independence.
