I have a doubt in binomial distribution.
Assume that a bag contains 5 black balls and 5 white balls. So the probability of drawing one black ball randomly is $\frac{1}{2}$. I want to find the probability of 6 black balls. Let the X be the number of black balls. So by binomial distribution $P(X=6)=\binom{10}{6}\big(\frac{1}{2}\big)^6\big(\frac{1}{2}\big)^{10-6}=0.205$.
But we had only 5 black balls. Probability of getting 6 black should have been $0$. Am I right? Please clarify this doubt.