If $ X $ ~ $ Hypergeometric (30,2,n) $, how much should $ n $ be worth $ P (X≥1) ≥0.5 $? The exercise

A batch of 30 bulbs contains two defective ones. A random sample with
  no batch return of size 4 is obtained.
a) What is the probability that the sample contains at least one
  defective bulb?
b) What should be the size of the sample, so that the probability of finding at least one defective bulb, is at least 0.5?

My solution
Let $X$ be the number of defective bulbs in a sample of size $n$, with
$$
 X \sim \mathrm{Hypergeometric}(30,2,n).
$$
a) $P(X≥1) = 1-f_X(0) = 0.252873563... \Leftrightarrow n=4$
b) $P(X≥1)≥0.5 \Leftrightarrow n=k$


*

*$P(X≥1) = 1-f_X(0) = 1-\dfrac{\binom{28}{n}}{\binom{30}{n}}≥0.5
   \Leftrightarrow n=k$

*$P(X≥1)$ grows as $n$ grows $\Rightarrow$ if $P(X≥1) = 0.5$ for $n = k \Rightarrow n≥k$ for which $P(X≥1)≥0.5$


And I came here, but I do not know if the reasoning is right. And if it's okay, I also do not know how to get the value of $k$, so that $n$ is greater than or equal to that $k$.
 A: This is a problem on the hypergeometric distribution. 
You may be able to unscramble the binomial coefficients and the
factorials to get an algebraic solution. The ratio of the binomial coefficients
simplifies considerably after several 'cancellations'.
But my guess is this may be a
drill problem in computing hypergeometric probabilities, and
that you are expected to use
the 'brute force' method of trying various values of $n$
until you get the answer.
That is feasible because the required
$n$ isn't very large.
You have already done the proper computation for $n = 4.$ In
R statistical software the computation to verify your answer is
as follows:
1 - dhyper(0, 2, 28, 4)
## 0.2528736

Now, it is easy to see the answer from the following, sightly
wasteful, computation:
n = 4:10;  pr = 1 - dhyper(0,2,28,n)
cbind(n, pr)      # 'cbind' binds together two column vectors to make table
  ##  n        pr
  ##  4 0.2528736
  ##  5 0.3103448
  ##  6 0.3655172
  ##  7 0.4183908
  ##  8 0.4689655
  ##  9 0.5172414
  ## 10 0.5632184

With a bit of strategy, you could vastly reduce the number of values
you'd have to try--either on a calculator or with MatLab, R, or some
other software.
A: Observe $$\begin{align*} \Pr[X \ge 1] &= 1 - \frac{\binom{28}{n}}{\binom{30}{n}} \\ 
&= 1 - \frac{28!}{n!(28-n)!} \cdot \frac{n!(30-n)!}{30!} \\
&= 1 - \frac{28!}{30!} \cdot \frac{(30-n)!}{(28-n)!} \\
&= 1 - \frac{28!}{(30)(29)(28!)} \cdot \frac{(30-n)(29-n)(28-n)!}{(28-n)!} \\
&= 1 - \frac{(30-n)(29-n)}{(30)(29)}. \end{align*}$$  Thus the requirement that $\Pr[X \ge 1] \ge \frac{1}{2}$ leads to the quadratic inequality in $n$  $$n^2 - 59n + 435 \le 0.$$  Solving for the roots using the quadratic formula gives $$\frac{59 - \sqrt{1741}}{2} \le n \le \frac{59 + \sqrt{1741}}{2},$$ from which we conclude that the least such integer $n$ that satisfies the criterion is $$n = 9.$$

It is worth noting that because there are only $2$ defective items, the solution can be found relatively easily.  In general, solving for the minimum $n$ involves finding the smallest real root of a degree $m$ polynomial where $m$ is the number of defects.  In cases where $m \ge 3$, it may be simpler to just perform trial computations and find the result by brute force.
