Using binomial approximation for calculating probability I am trying to solve the problem but stuck with 'at least' and 'at most', question is:
In a shipment of 20 engines, history shows that the probability of any one engine proving unsatisfactory is 0.1
a) Use the Binomial approximation to calculate the probability that more than 10 engines are defective?
In part a, more than 10 engines mean we are not including 10, I am writing it in R as below:
1 - pbinom(10, 20, .1)

Is it correct way of writing, do I need to add 10 or exclude 10?
b) Use the Poisson approximation to calculate the probability that at most three engines are defective?
again the same scenario is for part b, is it going to be ppois(3, 2) or we need to exclude 3 ?
 A: (a)  $X$ is the number of defective engines, $X \sim \mathsf{Binom}(n-20, p=.1).$
You seek $P(X > 10) = 1 - P(X \le 10) \approx 0.$ As you say this can be computed
in R as follows:
1 - pbinom(10, 20, .1)
[1] 7.088606e-07

x = 0:20;  PDF = dbinom(x, 20, .1)
plot(x, PDF, type="h", lwd=2, main="PDF of BINOM(20, .1)")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 abline(v = 10.5, col="red", lty = "dotted")


(b) Here again  $X \sim \mathsf{Binom}(n-20, p=.1),$
but you are using $Y \sim \mathsf{Pois}(\lambda = 2)$ as an approximation.
You seek $P(X \le 3) = 0.8670 \approx P(Y \le 3) = 0.8571.$ As you say the approximation
is computed in R as follows:
ppois(3, 2)
[1] 0.8571235

The exact binomial value is computed as follows:
pbinom(3, 20, .1)
[1] 0.8670467

The plot below shows the Binomial distribution (blue bars) and the
approximating Poisson distribution (brown). With either the exact binomial or its Poisson approximation, you want the sum of the heights of the bars to the left of the vertical dotted line.
pdf.p = dpois(x, 2)
hdr = "PDF of BINOM(20,.1) [blue] with Approximating POIS(2)"
plot(x-.05, PDF, type="h", col="blue", lwd=2, main=hdr)
 abline(h=0, col="green2");  abline(v=0, col="green2")
 lines(x+.05, pdf.p, type="h", col="brown")
 abline(v = 3.5, col="red", lwd=2, lty="dotted")


A: For your first question, for $n=20$ draws each with a probability of passing (in fact being unsatisfactory in your terms) $\epsilon=0.1$, the probability that more than $k=10$ pass (are unsatisfactory in your terms) is $P(>k|n, \epsilon) = 1 - P(<=k|n, \epsilon)$, ie the complement of the probability that at most $k$ fail.  That last number is the CDF of the binomial distribution at $k=10$.
The CDF at $k$ is given by R function $pbinom(k, n, \epsilon)$ (see https://www.tutorialspoint.com/r/r_binomial_distribution.htm), so your expression $1-pbinom(10, 20, 0.1)$ seems correct.
For your second question, I would assume the expected number of engine fails in 20 trials is $20 \times 0.1$, so I would calculate the Poisson CDF at 3 given the expectation of 2.
