Probability when choosing beads with repetitions There are 20 different beads, 4 of which are yellow. Choosing 30 beads with repetitions, I need to calculate:

*

*The probability that exactly 5 of the chosen beads are yellow

*The probability that the first and last beads are yellow and exactly 5 yellow beads were chosen (including the first and last beads)

The first one I calculated saying there are $4^5$ ways to chose 5 yellow beads and $16^{25}$ ways to choose the rest hence $$P_1 = \frac{4^5\cdot 16^{25}}{20^{30}}=\frac{4^{55}}{4^{30}\cdot5^{30}}=(0.8)^{25}\cdot(0.2)^{5}$$ As for the second probability I first tried choosing $\binom{5}{2}$ yellow beads for first and last and then multiply by 2 as they can change places and again multiply by the permutation of the remaining beads $28!$ hence $$P_2 = P_1\cdot\frac{\binom{5}{2}\cdot2\cdot28!}{30!}$$ but this is incorrect because there are repetitions. So the second attempt was choosing first and last yellow bead which can be done in $5^2$ and then $$P_2 = P_1\cdot\frac{5^2\cdot28!}{30!}$$ but this seems wrong too. My intuition says that the probability between 1 and 2 remains the same but I haven't found a way to prove (or disprove) it
 A: Because we are asked about probability, and not about the number of combinations, and because we are choosing with repetitions - we can ignore the identity of the individual beads and instead treat each choice as a $\frac{4}{20} = 20\%$ probability to choose yellow and $80\%$ probability to choose non-yellow.
For any given choice of the 5 places where yellow beads are chosen, the probability to choose it is  $$
0.2^5\cdot 0.8^{25}
= \frac{2^5\cdot 8^{25}}{10^5\cdot 10^{25}}
= \frac{2^5\cdot 2^{75}}{10^5\cdot 10^{25}}
= \frac{2^{80}}{10^{30}}
$$
Since each such case is disjoint to the others, we can calculate how many ways there are to choose these 5 places and multiply: $$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{30}{5}
= \frac{2^{80}}{10^{30}} \cdot \frac{30!}{5!\cdot 25!}
\approx 17.2279\%
}$$
For the second, the first and last places always have yellow beads so we only choose 3 places out of 28:
$$\boxed{
\frac{2^{80}}{10^{30}} \cdot \binom{28}{3}
= \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!}
\approx 0.396\%
}$$
A: I disagree with the posted solution.

*

*it is a Binomial $X\sim B(30;\frac{4}{20})$
Thus the solution is
$$\mathbb{P}[X=5]=\binom{30}{5}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 17.23\%$$
2.
$$\mathbb{P}[X=5, \text{ first and last Yellow}]=\binom{28}{3}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 0.40\%$$
This because, excluding the first and the last Yellow with probability $\Big(\frac{4}{20}\Big)^2$ the remaining $n=28$ draws can be represented by a binomial $Y\sim B(28;\frac{4}{20})$ and  you have to calculate
$\Big(\frac{4}{20}\Big)^2\times \mathbb{P}[Y=3]$
