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


2 Answers 2


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

  • 1
    $\begingroup$ There are 30 draws, not 20. $\endgroup$ Oct 20, 2020 at 13:03
  • $\begingroup$ @JaapScherphuis It was 20 before OP edited the question. Fixed. $\endgroup$
    – Idan Arye
    Oct 20, 2020 at 13:52
  • $\begingroup$ Sorry, I hadn't noticed the question had been edited. $\endgroup$ 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

$$\mathbb{P}[X=5]=\binom{30}{5}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 17.23\%$$


$$\mathbb{P}[X=5, \text{ first and last Yellow}]=\binom{28}{3}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 0.40\%$$

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

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


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