# 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:

1. The probability that exactly 5 of the chosen beads are yellow
2. 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

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: $$\boxed{ \frac{2^{80}}{10^{30}} \cdot \binom{28}{3} = \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!} \approx 0.396\% }$$

• There are 30 draws, not 20. Oct 20, 2020 at 13:03
• @JaapScherphuis It was 20 before OP edited the question. Fixed. Oct 20, 2020 at 13:52
• Sorry, I hadn't noticed the question had been edited. Oct 20, 2020 at 14:12

I disagree with the posted solution.

1. 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

$$\Big(\frac{4}{20}\Big)^2\times \mathbb{P}[Y=3]$$