Acceptance Sampling: Probability that the owner will return a shipment of fruits 
A grocery shop receives a monthly shipment of $1000$ fruits. The owner knows that usually $1 \%$ of the fruits are damaged when they arrive. To estimate the number of damaged ones, he takes a sample of $50$ fruits. If $2$ or more fruits in the sample are damaged then he returns the package.
Approximate the probability that the owner will return a shipment.

The way I've tried solving is:
$1 - \binom{50}{0}(0.99)^{50}(0.01)^{0} - \binom{50}{1}(0.99)^{49}(0.01)^{1} = 0.089$
Is this correct? They ask for an approximation so that's why I'm confused.
 A: Perhaps the invitation to 'approximate' has to do with a normal or Poisson approximation
to the binomial distribution. Approximation remains a popular topic, even though modern software
makes many approximations unnecessary.
Exact binomial probability. In your problem the number of damaged fruits $X$ seen when $n=50$ are
inspected has the distribution $\mathsf{Binom}(n = 50, p = .01).$
The probability $P(X \ge 2)$ of finding two or
more damaged fruits is can be found using the formula for the binomial
PDF (or PMF) as you have done.
$$P(X \ge 2) = 1 - P(X \le 1) = 0.0894,$$
so that less than 9% of the shipments will be refused.
In R statistical software, where pbinom is a binomial CDF and
dbinom is a binomial PDF, one obtains:
1 - pbinom(1, 50, .01)
## 0.08943531
sum(dbinom(2:50, 50, .01))
## 0.08943531

Also, many other statistical software packages and some statistical
calculators can be used to do similar computations.
Unsatisfactory normal approximation. Our random variable $X$ has $E(X) = np = 50(.01) = 0.5$ and a normal
approximation for a binomial random variable with a mean very near to 0 or 1
is not advisable. If one persisted in doing a normal approximation,
the 'best fitting' normal distribution is $\mathsf{Norm}(\mu =0.5, \sigma = 0.771)$ and one could standardize to get the the approximate probability
$0.097.$
Poisson approximation. In some cases a binomial distribution with mean $\mu$ is well approximated
by a Poisson distribution with the same mean $\mu.$ For your problem
$$P(X \ge 2) = 1 - P(X \le 1) = 1 - e^{-0.5}\left[\frac{0.5^0}{0!} + \frac{0.5^1}{1!}\right] = 0.0902.$$ 
On a calculator, the Poisson computation may be a little easier than the binomial one. In R:
 1 - ppois(1, .5)
 ## 0.09020401

Comparisons. Neither approximation is horrible. The Poisson approximation is 
noticeably better than the normal. But neither is as good as finding the
exact binomial probability.
The figure below shows the exact binomial probabilities along with the
approximate Poisson probabilities (centers of small circles), and the
'best fitting' normal density curve. The probability of interest is to
the right of the vertical dotted line.

