Question: In our project we have a population of 1000+ individuals. We picked a random sample of 107 individuals, but then we realized we needed more precision, so now we want to have a larger sample.

The problem is that sampling is expensive in our case, so we were thinking about picking another random sample of about 50 individuals from the remaining 900+ individuals, and adding it to the first sample in order to get a bigger sample without starting all over.

Would this new sample still be considered random?

**Edit:** We are taking this sample in order to estimate the distribution of the entire population. With the current sample, however, we are getting too much "holes" in our distribution, which leads to seemingly imprecise results.

Answer: 
**Correction**

I withdraw the previous answer (below) as wrong. @Glen\_b is correct, if the second sample is not based on any results in the first, then the overall probability of selection is $\dfrac{n}{N}$, the same as for a SRS (proof below). Further, you can treat the result as a SRS although the size was not fixed in advance.

However, you stated that you took the second sample because you required more precision. If you made the decision solely because standard errors for estimates were larger than expected, then bias in the final estimated standard errors will be small (Hansen, Hurwitz, and Madow (1953, pp.77-80) (provided that the initial SE themselves were precisely estimated). In that case, I would go ahead and treat the observations as having come from a SRS. If, however, you took the second sample to reduce a borderline p-value to a smaller value, then serious bias is possible.

**Proof that probability of selection is n/N:** 

Let $p\_1 =\dfrac{n\_1}{N}$ be the probability of selection in the first sample, and let $p\_2$ be the conditional probability of selection in the second sample: $p\_2 =\dfrac{n\_2}{N-n\_1}$. Let $n = n\_1 + n\_2$ be the final sample size. Then the probability of selection of an observation to the final sample is:

\begin{align}
p & = p\_1 + (1-p\_1)\thinspace p\_2 \\
& = \frac{n\_1}{N} + (1-\frac{n\_1}{N}) \frac{n\_2}{N-n\_1} \\
& = \frac{n\_1}{N} + \frac{N-n\_1}{N}\frac{n\_2}{N-n\_1} \\
& = \frac{n\_1 + n\_2}{N} \\
& = \frac{n}{N}
\end{align}

**Reference**: Hansen, MH, WN Hurwitz, and W Madow. 1953. Sample Survey Methods and Theory. Volume I Methods and Applications. New York: Wiley.

**Original Answer:**

This kind of problem frequently arises in practice and the solution is similar.

As @David Z implies, you must investigate the possibility of systematic differences between the first and second surveys and samples. 

The resulting combined sample can indeed be considered random, but not *simple random*. The probabilities of selection differ between the two samples, so the analysis will have to be weighted. You compute weights as follows:

Let the number in the population be $N$. Then the probability of selection for the original sample is:

$$
f\_1 = \frac{107}{N}
$$

To be selected in the second sample, one must first *not* be selected in the first sample; then be selected in the second: 

$$
f\_2 = (1 - f\_1) \frac{50}{N - 107}
$$

The sample weights will be $W = \dfrac{1}{f\_1}$ in sample 1 and $W = \dfrac{1}{f\_2}$ in sample 2.

To illustrate, suppose $N = 1020$, then the weights in the first and second sample will be $9.5327$ and $20.4$, respectively

These calculations will require modification if there was non-response in either sample.
I am not quite sure how random this sample could be considered. I don't have a great experience of statistics but the method you suggest doesn't sound very good to me. What I would do is put back the sample I took from the entire population, reorder them and take again a random sample from it. Or I would just take a new larger sample from the population and compare my results with the sample from the first try.

It might be a good idea to describe the limitations you have, as in why sampling is expensive for you, so other people can give a more elaborate answer.