The problem: First we state what we understand to be the problem. The probability that a randomly chosen item is defective is $0.01$. We take possibly very large samples, samples that are enormous by essentially all sampling standards. We want to find the probability that the number of defectives in the sample is $\le 5000$,
A comment mentions the binomial distribution. It is not clear that this is the appropriate distribution. If the number of defectives is exactly $10000$, then we are dealing with the hypergeometric distribution. However, for the kinds of calculations we are making, for all practical purposes it doesn't matter.
The normal approximation: Let random variable $X$ be the number of bads in a sample of size $n$. We want $\Pr(X\le 5000)$.
The random variable $X$ has mean $n(0.01)$ and variance $n(0.01)(0.99)$. So the standard deviation is $\sqrt{n}\sqrt{(0.01)(0.99)}$. For most practical purposes this is $\sqrt{n}/10$. (This number would require modification for another probability of a bad.)
The probability that $X\le 5000$ is extremely well approximated by
$$\Pr\left(Z\le \frac{5000-(0.01)n}{\sqrt{n}/10}\right),\tag{1}$$
where $Z$ is standard normal.
Some calculations: We do some calculations, for the various huge sample sizes mentioned in the OP.
$10$ percent: By (1), we want the probability that $Z\le 126.5$. This is $1$ for all practical and impractical purposes.
$30$ percent: By (1), we want the probability that $Z\le 36.5$. Again, this is $1$ for all practical purposes.
$40$ percent: Our probability is approximately the probability that $Z\le 15.8$. Again this is $1$ for all practical purposes.
$50$ percent: We want the probability that $Z\le 0$. This is $0.05$.
$60$ percent: We want the probability that $Z\le -15.8$. This is virtually equal to $0$.
Higher percentages give the same result, essentially $0$.
More calculations: The probability that $X\le 5000$ was virtually $1$ for the various percentages we calculated, up to but not including $50\%$, became $0.5$ at $50\%$, and had already dropped to virtually $0$ at $60\%$. We do some exploration of the fine stucture around $50\%$.
Look for example at $49\%$. Again by (1), the probability that $X\le 5000$ is approximately the probability that $Z\le 1.428$. This is about $0.92$, significantly away from $1$. By symmetry, with a sample size of $51\%$, we have $\Pr(X\le 5000)\approx 0.08$.
Finally, let's do the computation for $49.5\%$. We get $\Pr(X\le 5000)\approx 0.76$.
Remark: So interesting stuff happens only if we are looking at samples near the $50\%$ range. The phenomenon around $50\%$ is almost a "$0$-$1$ phenomenon, though on closer examination the transition turns out to be smooth.
Note that the normal approximation is not always appropriate for problems like this. For example, with the same probability $0.01$ of bad, To find the probability of exactly $3$ bad in a sample of $500$ I would suggest either the Poisson approximation to the binomial, or direct calculation of the binomial.
Generalities: We look at a more general situation. Let the population size be $N$, with $N$ large (in your case $N$ is $10^6$).
Let the probability of a bad be $p$, where $p$ is small (in your case $p=0.01$.) If $X$ is the number of bads in a sample of size $n$, then the standard deviation of $X$ is $\sqrt{np(1-p)}$.
The mean number of bads in the population is $pN$. Let $n=\alpha N$. We are looking at very large sample sizes $n$. The number $\alpha$ is the ratio $\frac{n}{N}$. For instance, if we are looking at a $40\%$ sample, then $\alpha=0.4$.
The expected number of bads in the population is $Np$. Your problem is to find the probability that $X\le \frac{1}{2}Np$. Note that
$$\Pr(X\le \frac{1}{2}Np)=\Pr\left(\frac{X-np}{\sqrt{np(1-p}}\le \frac{\frac{1}{2}Np-np}{\sqrt{np(1-p)}}\right).\tag{2}$$
The random variable $\frac{X-np}{\sqrt{np(1-p)}}$ is close to standard normal. Replace $n$ by $\alpha N$. After some simplification we find that we want
$$\Pr\left(Z \le \frac{1}{\sqrt{\alpha}\sqrt{p(1-p)}}\left(\frac{1}{2} -\alpha\right)\sqrt{N}\right).\tag{3}$$
We can do further simplification. For large $N$ and small $p$, the probability will be nearly one for $\alpha \lt 1/2$ but not too close to $1/2$, and nearly $0$ for $\alpha \gt 1/2$ but not too close to $1/2$. And $\sqrt{1-p}$ is close to $1$. So our probability is well approximated by
$$\Pr\left(Z\le\left(\frac{1}{2}-\alpha\right)\sqrt{2pN}\right).\tag{4}$$
This formula is easy to work with. As an example, from the normal tables we find that $\Pr(Z\le 4)\approx 0.999$. Let $p=0.01$ and $N=10^6$. Let's find out what $\alpha$ should be so that $\Pr(X\le 5000)\approx 0.999$. Calculation gives $\alpha=0.472$, about a $47\%$ sample.
Conclusion: Unless the sample size proportion is quite close to half the population size, the probability is for all practical purposes fully determined. There is not, however, a sudden shift at $50\%$. The shift is indeed rapid. With your numbers, for all practical purposes the only interesting interval is the one from about $47\%$ to $53\%$. For similar situations with different numbers (but $N$ still large, and $p$ small), Formula (4) should give very good quality estimates.
Perhaps the simplest general explanation of the phenomenon is that for $p$ of the size we have been looking at, or smaller, the variance of $X$ is relatively low. For $p=0.01$, it is about one-tenth of what the variance would be for $p=1/2$. Thus even for the $40\%$ case, $4000$ is a lot of standard deviation units away from $5000$.
It is not clear whether $Z$ is income or net income. No big deal, if we can handle one we can handle the other. We use the gross income interpretation.
Let $Z_1,Z_2,\dots, Z_{10}$ be the amount of money made from digs $1,2,\dots,10$. Then $Z=Z_1+Z_2+\cdots+Z_{10}$. By the linearity of expectation, we have $E(Z)=E(Z_1)+\cdots +E(Z_{10})=10E(Z_1)$.
To find $E(Z_1)$, note that $Z_1=0$ with probability $1-p$, where $p$ is the probability of finding oil if one digs, currently unreadable. And given that the well was successful, the expectation is $50000$. Thus $E(Z_1)=(1-p)(0)+(p)(50000)$.
For the probability that $Z\gt 10000$ given $Y=1$, we just want the probability that an exponential with mean $50000$ is greater than $10000$.
Remark: If we interpret $Z$ as net income, for the expectation question subtract $100000$.
For the probability question, find the probability that an exponential with mean $50000$ is greater than $110000$.
Best Answer
For any $k$ such that $1\leq k\leq n-1$:
\begin{align} P(Y=k) &= \sum_{j=0}^{\infty} P(X=k+jn) \\ &= \sum_{j=0}^{\infty} q^{k+jn-1}p \\ &= q^{k-1}p \sum_{j=0}^{\infty} \left(q^{n}\right)^j \\ &= \dfrac{q^{k-1}p}{1-q^{n}} = \dfrac{P(X=k)}{1-q^{n}}. \\ \end{align}
Also, the special case of $Y=0$ since $X=0$ can't occur:
\begin{align} P(Y=0) &= \sum_{j=1}^{\infty} P(X=jn) \\ &= \sum_{j=1}^{\infty} q^{jn-1}p \\ &= q^{n-1}p \sum_{j=0}^{\infty} \left(q^{n}\right)^j \\ &= \dfrac{q^{n-1}p}{1-q^{n}} = \dfrac{P(X=n)}{1-q^{n}}. \\ \end{align}
Note that for large $n$, and $p$ not near $0$, the probability distribution of $Y$ is almost the same as that of $X$. For your last question, the effect will depend on the distribution that $X$ has.