Answer:
With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.
Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.
Step-by-step explanation:
For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:

In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
is the number of different combinations of x objects from a set of n elements, given by the following formula.

Sampling with replacement:
I consider a success choosing a black ball, so 
We want 2 black balls and 2 white, 2 + 2 = 4, so
, and we want P(X = 2).


With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.
Sampling without replacement:
150 + 50 = 200 total balls, so 
Sample of 4, so 
50 are black, so 
We want P(X = 2).

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.