I am solving a problem similar to this:
In a sample of 100 balls there are 33 red balls and 67 blue balls. What is the probability of picking randomly $X$ balls and have less than half red balls
I was just introduced to hypergeometric distirbutions and my solution was something like this:
(Imagining that $X$ = 13 and $K$ = number of red balls)
$P( K \leq 6) = \displaystyle\sum_{n = 0}^{6}\frac{{33 \choose n}{67 \choose 13-n}}{{100 \choose 13}} \approx 91,63\%$
My problem (if my solution has no problem) is that as i am increasing the sample size, i notice a weird pattern in my probabilities of finding less than half of red balls (only the sample size increases, the number of balls for each color is the same)
This is the results i'm getting
$P(X = 14, K \leq 6) \approx 87,43\%$
$P(X = 15, K \leq 7) \approx 93,29\%$
$P(X = 16, K \leq 7) \approx 89,93\%$
Why everytime i increase my sample size but my $K$ remains the same the probability of having less than half of red balls lowers? Shouldn't it increase because i have a higher number of blue balls?
There is nothing wrong with your answer, but your observation that the probability should increase as the sample size increases is incorrect.
Remember, we are computing cumulative probabilities. To understand what is happening, let us analyse the first two cases, computing probabilities directly.
With sample size $13, P(0\; red) = \frac{67}{100}\frac{66}{99}...\frac{55}{88}$
wheras for sample size $14$, we need one more multiplier, $\frac{54}{87}=\frac{18}{29}$
On the other hand, for $P(6\;red)$ the multiplier for sample size $14$ taking into account changes in permutations will be $\frac{60}{87}\frac{14}{8} = \frac{35}{29}$