This is quite an interesting question, since it involves probability pertaining to overlapping random sets. I'm going to solve this for a more general problem where we have $K$ servers. The solution turns out to relate to the maximum-count statistic in a multinomial distribution. By my calculations, the probability of interest is approximately $0.45214$ ---i.e., you have a little over a 45% chance that at least half the serveres will run Test B simultaneously at some point in the hour.
Simplify the problem: To simplify the problem, let's break the hour down into 5-minute blocks, and denote these blocks of time as $t=1,...,12$. Denote the set of time blocks taken up by test B as $\mathscr{B}$. Based on the stipulated times for the tests, here are the possible outcomes for test B based on the test order, and the number of ways of getting that outcome (for convenience, I have ordered these cases by the lower bound of the set $\mathscr{B}$).
$$\begin{matrix} \text{Test order} & & & & \text{Result of Test }B & & & & \text{Number of ways} \\[6pt] B,... & & & & \mathscr{B} = \{ 1, 2 \} & & & & 6 & & & & \text{*} \\[6pt] A,B,... & & & & \mathscr{B} = \{ 2, 3 \} & & & & 2 & & & & \text{*} \\[6pt] C,B,... & & & & \mathscr{B} = \{ 4, 5 \} & & & & 2 & & & & \text{**} \\[6pt] ...,B,D & & & & \mathscr{B} = \{ 5, 6 \} & & & & 2 & & & & \text{**} \\[6pt] D,B,... & & & & \mathscr{B} = \{ 7, 8 \} & & & & 2 & & & & \text{***} \\[6pt] ...,B,C & & & & \mathscr{B} = \{ 8, 9 \} & & & & 2 & & & & \text{***} \\[6pt] ...,B,A & & & & \mathscr{B} = \{ 10, 11 \} & & & & 2 & & & & \text{****} \\[6pt] ...,B & & & & \mathscr{B} = \{ 11, 12 \} & & & & 6 & & & & \text{****} \\[6pt] \end{matrix}$$
(Note that there are $4! = 24$ total cases that can occur from the ordering of the four tests. This matches the total of the last column shown here.) Now, let's look at the overlapping cases. We can see that there are pairs of overlapping cases shown by the stars in the table. So, the total highest number of servers running Test B at a given time is going to be equal to the highest number of servers that have an outcome within one of the stipulated pairs of outcomes for the test order.
We can see from the above that all that matters for the purposes of our question are the four groups of outcomes denoted by the stars. So, let us now denote the test outcome for a single server by four possibilities $X = 1, 2, 3, 4$, corresponding to the number of stars in the table above. Using the number of ways of getting the test orders in the table, we have the probabilities:
$$\begin{align} \mathbb{P}(X=1) &= \frac{6+2}{24} = \frac{8}{24} = \frac{1}{3}, \\[12pt] \mathbb{P}(X=2) &= \frac{2+2}{24} = \frac{4}{24} = \frac{1}{6}, \\[12pt] \mathbb{P}(X=3) &= \frac{2+2}{24} = \frac{4}{24} = \frac{1}{6}, \\[12pt] \mathbb{P}(X=4) &= \frac{6+2}{24} = \frac{8}{24} = \frac{1}{3}. \\[12pt] \end{align}$$
For a single server, the categorical random variable $X$ encapsulates all that really matters in terms of the test-order. If two different servers have the same value for this random variable then they are both running Test B at some point in time, and if two different servers have a different value for this random variable then they never run Test B at the same time.
Solve the problem: Now that we have established the random variable of interest for a single server, we can solve the problem. Suppose we have $K$ servers and let the test-order outcomes for these servers be denoted by the categorical random variables:
$$X_1, ..., X_K \sim \text{IID Categorical}(\mathbf{p}) \quad \quad \quad \mathbf{p} = (\tfrac{1}{3}, \tfrac{1}{6}, \tfrac{1}{6}, \tfrac{1}{3}).$$
Letting $N_{x, K} \equiv \sum_{i=1}^K \mathbb{I}(X_i = x)$ denote the counts for outcomes $x=1,2,3,4$, the count vector for all servers and test-orders is a multinomial random vector:
$$\mathbf{N}_K = (N_{1, K}, N_{2, K}, N_{3, K}, N_{4, K}) \sim \text{Mu}(K, \mathbf{p}),$$
and the highest number of servers that run Test B at any given time is the maximum-count value:
$$M_K \equiv \max \{ N_{1, K}, N_{2, K}, N_{3, K}, N_{4, K} \}.$$
The probability of interest in this question is $\mathbb{P}(M_K \geqslant K/2)$, which is the probability that Test B runs on at least half the servers at some point in the hour. This is something you can obtain from the distribution of the maximum-count statistic. This distribution is fairly complicated, and it is usually approximated by an asymptotic form such as the Gumbel distribution. However, for small values (as in the present case) it can be simulated reasonably easily, and this is generally how you would proceed from this point.
Simulate the distribution: It is fairly simple to get a simulated estimate for the distribution of the maximum-count statistic using R. In the code below I will simulate ten-million outcomes for the case with $K=10$ servers (as in your problem). The simulated probability of having Test B running on at least half the servers simultaneously is $\mathbb{P}(M_{10} \geqslant 5) \approx 0.45214$.
#Set the parameters
K <- 10
SIMS <- 1e7
#Set up the count matrix
COUNTS <- matrix(0, nrow = SIMS, ncol = 5)
rownames(COUNTS) <- sprintf('Simulation[%s]', 1:SIMS)
colnames(COUNTS) <- c(sprintf('Outcome[%s]', 1:4), 'MaxCount')
#Generate a large number of count vectors
set.seed(418184220)
for (i in 1:SIMS) {
SAMPLE <- sample.int(4, size = K, replace = TRUE, prob = c(2, 1, 1, 2)/6)
for (x in 1:4) { COUNTS[i, x] <- sum(SAMPLE == x) }
COUNTS[i, 5] <- max(COUNTS[i, 1:4]) }
#Compute approximate distribution
DIST <- rep(0, K+1)
names(DIST) <- sprintf('MaxCount[%s]', 0:K)
for (M in 0:K) { DIST[M+1] <- sum(COUNTS[, 5] == M)/SIMS }
#Plot the simulated distribution
barplot(DIST, names.arg = 0:K, ylim = c(0, 0.5), col = 'blue',
main = 'Simulated MaxCount Distribution',
xlab = 'Maximum-Count', ylab = 'Probability')
#Compute the probability of interest
sum(DIST[6:11])
[1] 0.45214
