Set of white/red balls, select three, get the color of two of those You have a (big) bag with twelve white and four red balls. So, chances of getting a red ball when selecting a random ball from the big bag is (4/16) 25%.
Now you select three random balls out of the big bag bag without looking. En put those three in a (small) second bag. From the second bag you take two random balls, they are both white, and you put them back into the small bag.
What's the chance of a ball in the big bag being red, and what are the chances of the balls in the small bag being red?
 A: An approach to solve this is to tabulate the potential outcomes
Small bag contents   Observation of 2 whites    Probability
0 red                Yes                        (55/140) * 1
0 red                No                         (55/140) * 0
1 red                Yes                        (66/140) * (1/3) 
1 red                No                         (66/140) * (2/3)
2 red                Yes                        (18/140) * 0
2 red                No                         (18/140) * 1
3 red                Yes                        ( 1/140) * 0                       
3 red                No                         ( 1/140) * 1

The probabilities in the right column are computed as the product of two hypergeometric distributions
$$P(\text{$r$ red and two white}) = \underbrace{\frac{{R\choose r}{W\choose n-r}}{N\choose n}}_{\substack{\text{probability $r$ red} \\ \text{in small bag}}} \cdot \underbrace{\frac{{r\choose n-2}{n-r\choose 2}}{n\choose 2}}_{\substack{\text{probability 2 white} \\ \text{given $r$ red in small bag}}} $$
Where $R$ and $W$ are the number of red and white balls in the big bag, and $n$ is the number of balls that are placed in the small bag.
Then you can compute the conditional probability for $x$ red balls in the bag by Bayes rule:
$$P(x \text{ red in small bag} | \text{given 2 whites observed}) = \frac{P(x \text{ red in small bag and 2 whites observed})}{P(\text{2 whites observed})}$$
So we have
$$P(x) = \begin{cases} \frac{55/140}{22/140 + 55/140} =\frac{5}{7}    &\quad {\text{if $x=0$}} \\
\frac{22/140}{22/140 + 55/140} =\frac{2}{7} &\quad {\text{if $x=1$}} \\
0 &\quad {\text{if $x=2$}} \\
0 &\quad {\text{if $x=3$}} \\
\end{cases}$$
Simulation
With a computer code you can simulate this. The result is 2242 out of 14000 cases with 1 red ball in the small bag and drawing 2 whites, and 5515 out of 14000 cases with 0 red balls in the small bag and drawing 2 whites
n_reds = c() ### a vector where we store our results

### we do the following a 14 000 times
set.seed(1)

for (i in 1:14000) {
  ### a vector with 4 ones (red) and 12 zeros (white) representing the balls 
  balls <- c(rep(1,4),rep(0,12))
  
  ### draw a sample for the small bag
  smallbag <- sample(balls, 3, replace = FALSE)
  
  ### draw a sample for the hand
  hand <- sample(smallbag, 2, replace = FALSE)
  
  ### if no reds in the hand
  if (sum(hand) == 0) {
    ### the count the reds in the smallbag
    reds = sum(smallbag)
    ### add them to the results 
    n_reds = c(n_reds,reds)
  }
}


### number of cases
no1 <- sum(n_reds == 1)
no0 <- sum(n_reds == 0)

no1/10^4
no0/10^4

