What is the probability that both balls you draw a red Assuming you do not replace the first red ball before drawing the second one?

I'm currently leaning probability and I came out with this problem, I can't find a solution

Lisa Williams has correctly calculated the probability. As she observed, the probability of selecting a red ball on the first draw is $11/16$ since $11$ of the original $16$ balls are red. Since we are drawing without replacement, the probability of selecting a second red ball is $10/15$ since $10$ of the $15$ remaining balls are red. Thus, the probability of selecting two red balls without replacement is $$P(\text{two red}) = P(\text{red})P(\text{red} \mid \text{first ball is red}) = \frac{11}{16} \cdot \frac{10}{15} = \frac{11}{24}$$ Observe that the probability that the second ball selected is red depends on the fact that the first ball was red.

The condition that $$P(F \mid E) = \frac{P(E \cap F)}{P(E)} = \frac{P(E)P(F)}{P(E)} = P(F)$$ means that events $E$ and $F$ are independent since the probability that event $F$ occurs given that event $E$ has occurred is equal to the probability that event $F$ occurs. That is not the case here since the probability that the second ball selected is red depends on whether the first ball selected was red, so $P(\text{red} \mid \text{first ball is red}) \neq P(\text{red})P(\text{red})$. It would be the case if the balls were replaced after each draw.

An alternative method of calculating the probability of selecting two red balls without replacement is $$P(\text{both red}) = \frac{\binom{11}{2}}{\binom{16}{2}}$$ where the numerator represents the number of ways of selecting two of the eleven red balls in the urn and the denominator represents the number of ways of selecting two of the sixteen balls in the urn.