Solving cumulative distribution: difference between with and without replacement Here is the problem that I am to solve:
Draw a sample of 5 balls from a box containing 5 red and 5 black balls. What is the probability that your sample will contain at least 3 red balls? Show work for both with and without replacement.
I understand that with replacement is a binomial distributive function:
$$p(x\ge3)=C_3^5(0.5)^3(0.5)^2+C_4^5(0.5)^4(0.5)^1+C_5^5(0.5)^5(0.5)^0=0.5$$
Perhaps this shows where my lack of knowledge in probability shows, but how does this change for without replacement?
 A: With replacement, the probability that you draw at least three red balls is the probability that you draw three red and two black, plus the probability that you draw four red and one black, plus the probability that you draw all five red balls.  That is,
$$
P(X \geq 3) = \frac{\binom{5}{3}\binom{5}{2}+\binom{5}{4}\binom{5}{1}+\binom{5}{5}\binom{5}{0}}{\binom{10}{5}} = \frac{100+25+1}{252} = \frac{1}{2}
$$
The result is the same, but the derivation is different.  The answer in each case has to be $1/2$ because it must be the same as the probability that at least three black balls are drawn—by symmetry—and those two cases are mutually disjoint and cover the entire range of possibilities.
The answers would not be the same if you were the consider the probability that at least four red balls are drawn.  With replacement, we have a probability of $6/32 = 0.1875$; without replacement, we have a probability of $26/252 \doteq 0.10317$.  The case of at least five red balls being chosen is even easier to check by inspection: With replacement, it's just $(1/2)^5 = 1/32 = 0.03125$, while without replacement, there's only one possibility out of $252$ possible draws, yielding $1/252 \doteq 0.0039683$.
